This document contains 35 questions related to data structures and algorithms. It covers topics like data structures used in different areas like databases, networks and hierarchies. Other topics covered include trees, graphs, sorting, hashing and file structures. Sample problems are given related to these topics to test understanding.
Data Structure is a way of collecting and organising data in such a way that we can perform operations on these data in an effective way. Data Structures is about rendering data elements in terms of some relationship, for better organization and storage. For example, we have data player's name "Virat" and age 26. Here "Virat" is of String data type and 26 is of integer data type.
We can organize this data as a record like Player record. Now we can collect and store player's records in a file or database as a data structure. For example: "Dhoni" 30, "Gambhir" 31, "Sehwag" 33
In simple language, Data Structures are structures programmed to store ordered data, so that various operations can be performed on it easily.
The document provides 29 sample questions on data structures and algorithms along with their answers. Some of the key questions covered include:
1. What is a data structure and examples of areas where they are applied extensively such as compiler design, operating systems, etc.
2. Major data structures used in relational databases, network and hierarchical data models.
3. Data structures used to perform recursion and evaluate arithmetic expressions.
4. Sorting algorithms like quicksort illustrated through an example.
5. Properties of different trees including the number of possible trees with a given number of nodes and number of null branches in a binary tree.
The summary hits the main topics covered in the document such as common
Stacks, Queues, Binary Search Trees - Lecture 1 - Advanced Data StructuresAmrinder Arora
This document introduces the course CS 6213 - Advanced Data Structures. It discusses what data structures are, how they are designed to efficiently support specific operations, and provides examples. Common data structures like stacks, queues, linked lists, trees, and graphs are introduced along with their basic operations and implementations. Real-world applications of these data structures are also mentioned.
The document discusses topics related to trees and graphs. It covers definitions and concepts of trees, binary trees and their representations. It discusses operations on binary trees like tree traversals. Binary search trees and heap trees are also covered. The document also provides an introduction to graphs and discusses graph representations and traversals. It concludes with the topic of minimal spanning trees.
2. Linear Data Structure Using Arrays - Data Structures using C++ by Varsha P...widespreadpromotion
This document discusses arrays and ordered lists as linear data structures. It describes arrays as a collection of homogeneous data elements that allow direct access via indexes. The key properties of arrays include sequential storage of elements, random access, and calculation of element addresses based on base address and offsets. The document also covers array implementation in C++, one-dimensional and two-dimensional arrays, row-major and column-major representations, and the use of arrays to represent polynomials, strings and sparse matrices. Finally, it provides a brief introduction to ordered lists and their representation as linear data structures.
The following presentation consists of information about the application of matrices. The ppt particularly focuses on the its use in cryptography i.e. encoding and decoding of messages.
1) Introduction to Trees.
2) Basic terminologies
3) Binary tree
4) Binary tree types
5) Binary tree representation
6) Binary search tree
7) Creation of a binary tree
8) Operations on binary search tree Trees
Data Structure is a way of collecting and organising data in such a way that we can perform operations on these data in an effective way. Data Structures is about rendering data elements in terms of some relationship, for better organization and storage. For example, we have data player's name "Virat" and age 26. Here "Virat" is of String data type and 26 is of integer data type.
We can organize this data as a record like Player record. Now we can collect and store player's records in a file or database as a data structure. For example: "Dhoni" 30, "Gambhir" 31, "Sehwag" 33
In simple language, Data Structures are structures programmed to store ordered data, so that various operations can be performed on it easily.
The document provides 29 sample questions on data structures and algorithms along with their answers. Some of the key questions covered include:
1. What is a data structure and examples of areas where they are applied extensively such as compiler design, operating systems, etc.
2. Major data structures used in relational databases, network and hierarchical data models.
3. Data structures used to perform recursion and evaluate arithmetic expressions.
4. Sorting algorithms like quicksort illustrated through an example.
5. Properties of different trees including the number of possible trees with a given number of nodes and number of null branches in a binary tree.
The summary hits the main topics covered in the document such as common
Stacks, Queues, Binary Search Trees - Lecture 1 - Advanced Data StructuresAmrinder Arora
This document introduces the course CS 6213 - Advanced Data Structures. It discusses what data structures are, how they are designed to efficiently support specific operations, and provides examples. Common data structures like stacks, queues, linked lists, trees, and graphs are introduced along with their basic operations and implementations. Real-world applications of these data structures are also mentioned.
The document discusses topics related to trees and graphs. It covers definitions and concepts of trees, binary trees and their representations. It discusses operations on binary trees like tree traversals. Binary search trees and heap trees are also covered. The document also provides an introduction to graphs and discusses graph representations and traversals. It concludes with the topic of minimal spanning trees.
2. Linear Data Structure Using Arrays - Data Structures using C++ by Varsha P...widespreadpromotion
This document discusses arrays and ordered lists as linear data structures. It describes arrays as a collection of homogeneous data elements that allow direct access via indexes. The key properties of arrays include sequential storage of elements, random access, and calculation of element addresses based on base address and offsets. The document also covers array implementation in C++, one-dimensional and two-dimensional arrays, row-major and column-major representations, and the use of arrays to represent polynomials, strings and sparse matrices. Finally, it provides a brief introduction to ordered lists and their representation as linear data structures.
The following presentation consists of information about the application of matrices. The ppt particularly focuses on the its use in cryptography i.e. encoding and decoding of messages.
1) Introduction to Trees.
2) Basic terminologies
3) Binary tree
4) Binary tree types
5) Binary tree representation
6) Binary search tree
7) Creation of a binary tree
8) Operations on binary search tree Trees
This document discusses binary search trees and AVL trees. It begins with an introduction to binary search trees and their properties. It then describes AVL trees, which are self-balancing binary search trees that ensure the height difference between any two subtrees of a node is at most 1. The document provides pseudocode for performing basic operations on AVL trees like insertion, single rotations, double rotations, and calculating the balancing factor. It also includes examples of calculating heights and balancing factors for AVL tree nodes.
Graphs are one of the most important non-linear data structures that represent relationships between elements. Graphs can be classified as directed or undirected. Common graph representations include adjacency lists and matrices. Graph algorithms like depth-first search, breadth-first search, minimum spanning trees (using Prim's and Kruskal's algorithms), and shortest paths (using Dijkstra's algorithm) are used to traverse and analyze graph structures.
Engineering Research Publication
Best International Journals, High Impact Journals,
International Journal of Engineering & Technical Research
ISSN : 2321-0869 (O) 2454-4698 (P)
www.erpublication.org
This document discusses graphs and networks. It defines key terms related to graphs like nodes, branches, trees, and loops. It explains properties of trees, including that a tree with n nodes will have n-1 branches and no closed paths. It also discusses incidence matrices, cut-set matrices, and loop matrices which relate to representing graphs and networks mathematically through matrices. Formulas are provided for relationships between the number of trees, branches, and nodes in a graph.
This document discusses stacks and queues as data structures. It begins by explaining what a stack is, noting that a stack follows last-in, first-out ordering. It then provides an analogy using mail delivery to explain the stack concept. The document goes on to provide Java code examples for implementing a stack. It also gives examples of using a stack to reverse a word and check balanced parentheses. Next, the document defines queues as first-in, first-out data structures and provides Java code for implementing a queue. It concludes by explaining how stacks can be used to parse arithmetic expressions by first converting them to postfix notation.
This document discusses affine array accesses in compiler design. It defines an affine access as one where the array index is expressed as an affine expression of loop indexes and constants. Affine accesses can be represented as a matrix-vector calculation that maps the iteration space to data space. Examples are given of affine accesses and how they can be written as tuples representing the mapping between indexes and array elements. Non-affine accesses are also discussed, such as those involving sparse matrices. Exercises are provided to represent given array accesses as affine tuples.
This document discusses data structures and their implementation in C++. It begins by defining the objectives of understanding data structures, their types, and operations. It then defines data and data structures, and describes how data is represented in computer memory. The document classifies data structures as primitive and non-primitive, and describes common operations on each. It provides examples of linear and non-linear data structures like arrays, stacks, queues, and trees. The document concludes by explaining arrays in more detail, including their representation in memory and basic operations like traversing, searching, and sorting.
This document discusses various network topology concepts including nodes, branches, loops, trees, and different matrix representations of networks. It defines key terms like nodes, branches, loops, meshes, and oriented graphs. It also describes tree concepts such as twigs, links, and co-trees. Finally, it discusses different matrix representations of networks including the incidence matrix, loop matrix, tie-set matrix, cut-set matrix, and formulations of network equilibrium equations in node, mesh, and cut-set forms.
Data structure using c bcse 3102 pcs 1002SANTOSH RATH
This document contains a data structures exam from 2008 with 8 pages. It includes 10 multiple choice questions covering topics like data structures, binary trees, queues, stacks, graphs, and sorting algorithms. It also contains 8 additional problems requiring algorithms and programs related to linked lists, binary trees, heaps, graphs, searching, and polynomial representations. Students are instructed to answer question 1 and 5 of the additional problems.
Encryption and decryption are both methods used to ensure the secure passing of messages and other sensitive documents and information. The encryption process plays a major factor in our technology advanced lives. Encryption basically means to convert the message into code or scrambled form. Advanced Encryption Standard (AES) is a specification for the encryption of electronic data. It has been adopted by the U.S. government and is now used worldwide. AES is a symmetric-key algorithm, meaning the same key is used for both encrypting and decrypting the data. This paper defines the method to enhance the block and key length of the conventional AES.
This document discusses different types of balanced binary search trees including AVL trees, B-trees, and B+-trees. It provides examples of how to construct and perform operations on each type of tree. AVL trees balance after each insertion or deletion by performing tree rotations if needed. B-trees allow nodes to have more than two children and remain balanced by splitting full nodes. B+-trees are like B-trees but also allow sequential record traversal through leaf nodes. Threaded binary trees add threads to leaf nodes for efficient in-order traversal.
The document discusses binary search trees and their properties. It explains that a binary search tree is a binary tree where every node's left subtree contains values less than the node's value and the right subtree contains greater values. Operations like search, insert, delete can be done in O(h) time where h is the height of the tree. The height is O(log n) for balanced trees but can be O(n) for unbalanced trees. The document also provides examples of using a binary search tree to sort a set of numbers in O(n log n) time by building the BST and doing an inorder traversal.
Application of Matrices in real life | Matrices application | The MatricesSahilJhajharia
Matrices can be used to transform vectors by changing their magnitude and direction. Matrices are useful for applications like rotating vectors, solving systems of linear equations, and encoding/decoding messages for cryptography. As an example, a message can be encoded using a matrix, transmitted as encoded values, and then decoded by the receiver using the inverse matrix. Eigenvectors are vectors that only change in magnitude and not direction when multiplied by a matrix. They can be used to model real-world systems changing over time, like populations of humans and zombies.
Presentation given at DMZ about Data Structure Graphs.
Also known as Applying Social Network Analysis Techniques to Data Modeling and Data Architecture
OVERVIEW:
Introduction
Definition
Example of Threaded BT.
Types & Structure
One-way .
Double-way.
Structure.
Traversal
Algorithm for Traversal
Traversal Example
Inserting
Algorithm for Inserting
Inserting Example
Comparison With Binary Tree
Advantages and Disadvantages
Why Threaded BT are used?
Conclusion
Reference
The document discusses various data addressing modes used in microprocessors, including register, immediate, direct, indirect, base-plus-index, register relative, base relative-plus-index, and scaled-index addressing. It also covers program addressing modes like direct, relative, and indirect jumping and calling. Finally, it discusses stack addressing and how the push and pop instructions are used to place data onto and remove data from the stack.
Cryptography an application of vectors and matricesdianasc04
This document discusses cryptography and various encryption methods using matrices. It introduces shift ciphers, stretch ciphers, combination ciphers, and the Vigenere cipher which uses a keyword to shift between cipher alphabets. It provides examples of encoding and decoding messages with these ciphers and discusses how matrices can represent and manipulate encrypted data. It also considers the benefits and limitations of different encryption methods and how cryptography applies to fields like warfare.
This document provides an introduction and overview of a discrete mathematics course. It explains that discrete mathematics lays the mathematical foundations for many computer science topics. It lists common topics covered in the course like logic, sets, functions, counting, recursion, graphs and trees. It emphasizes that discrete mathematics deals with discrete, distinct objects like those used in computing. Mastering this subject involves actively practicing the concepts rather than just hearing or reading about them.
1) The document is an assignment submission on data structures covering queues, linked lists, and trees.
2) It discusses the concepts of queues including FIFO operations, applications such as call centers, and implementations using arrays or linked lists.
3) Linked lists are described as dynamic structures that connect nodes through pointers and are used for stacks, queues, and graphs. Common types include singly, doubly, and circular linked lists.
4) Trees are hierarchical structures where each node has a maximum of two children. Binary trees, their terminology, types such as full and balanced binary trees, and advantages are outlined.
O'Donnell Consulting Engineers evaluated vibrations in the framework supporting the Hubble Space Telescope. Tests and analyses showed the vibrations were caused by shortening of transverse bracing members. The company added lateral restraints to the bracing, easily solving the vibration problems. O'Donnell provides engineering design, analysis, materials consulting, and problem solving services to clients worldwide, with experience spanning industries and challenging issues like fabrication, fatigue, corrosion, and failure analysis. For over 40 years, the company has taken on complex technical problems for clients in various sectors.
The document summarizes the design process of Team RAPTOR for the 2015 NASA Robotic Mining Competition. It describes the systems engineering approach used, including defining requirements, developing a concept of operations, and conducting various design reviews. Key aspects of the preliminary and final robot design are discussed, such as using magnesium alloys for the structure to reduce weight and implementing a conveyor system for regolith transport to minimize tipping risks. The team's goal was to develop an autonomous robot called DAIR capable of deep icy regolith excavation.
This document discusses binary search trees and AVL trees. It begins with an introduction to binary search trees and their properties. It then describes AVL trees, which are self-balancing binary search trees that ensure the height difference between any two subtrees of a node is at most 1. The document provides pseudocode for performing basic operations on AVL trees like insertion, single rotations, double rotations, and calculating the balancing factor. It also includes examples of calculating heights and balancing factors for AVL tree nodes.
Graphs are one of the most important non-linear data structures that represent relationships between elements. Graphs can be classified as directed or undirected. Common graph representations include adjacency lists and matrices. Graph algorithms like depth-first search, breadth-first search, minimum spanning trees (using Prim's and Kruskal's algorithms), and shortest paths (using Dijkstra's algorithm) are used to traverse and analyze graph structures.
Engineering Research Publication
Best International Journals, High Impact Journals,
International Journal of Engineering & Technical Research
ISSN : 2321-0869 (O) 2454-4698 (P)
www.erpublication.org
This document discusses graphs and networks. It defines key terms related to graphs like nodes, branches, trees, and loops. It explains properties of trees, including that a tree with n nodes will have n-1 branches and no closed paths. It also discusses incidence matrices, cut-set matrices, and loop matrices which relate to representing graphs and networks mathematically through matrices. Formulas are provided for relationships between the number of trees, branches, and nodes in a graph.
This document discusses stacks and queues as data structures. It begins by explaining what a stack is, noting that a stack follows last-in, first-out ordering. It then provides an analogy using mail delivery to explain the stack concept. The document goes on to provide Java code examples for implementing a stack. It also gives examples of using a stack to reverse a word and check balanced parentheses. Next, the document defines queues as first-in, first-out data structures and provides Java code for implementing a queue. It concludes by explaining how stacks can be used to parse arithmetic expressions by first converting them to postfix notation.
This document discusses affine array accesses in compiler design. It defines an affine access as one where the array index is expressed as an affine expression of loop indexes and constants. Affine accesses can be represented as a matrix-vector calculation that maps the iteration space to data space. Examples are given of affine accesses and how they can be written as tuples representing the mapping between indexes and array elements. Non-affine accesses are also discussed, such as those involving sparse matrices. Exercises are provided to represent given array accesses as affine tuples.
This document discusses data structures and their implementation in C++. It begins by defining the objectives of understanding data structures, their types, and operations. It then defines data and data structures, and describes how data is represented in computer memory. The document classifies data structures as primitive and non-primitive, and describes common operations on each. It provides examples of linear and non-linear data structures like arrays, stacks, queues, and trees. The document concludes by explaining arrays in more detail, including their representation in memory and basic operations like traversing, searching, and sorting.
This document discusses various network topology concepts including nodes, branches, loops, trees, and different matrix representations of networks. It defines key terms like nodes, branches, loops, meshes, and oriented graphs. It also describes tree concepts such as twigs, links, and co-trees. Finally, it discusses different matrix representations of networks including the incidence matrix, loop matrix, tie-set matrix, cut-set matrix, and formulations of network equilibrium equations in node, mesh, and cut-set forms.
Data structure using c bcse 3102 pcs 1002SANTOSH RATH
This document contains a data structures exam from 2008 with 8 pages. It includes 10 multiple choice questions covering topics like data structures, binary trees, queues, stacks, graphs, and sorting algorithms. It also contains 8 additional problems requiring algorithms and programs related to linked lists, binary trees, heaps, graphs, searching, and polynomial representations. Students are instructed to answer question 1 and 5 of the additional problems.
Encryption and decryption are both methods used to ensure the secure passing of messages and other sensitive documents and information. The encryption process plays a major factor in our technology advanced lives. Encryption basically means to convert the message into code or scrambled form. Advanced Encryption Standard (AES) is a specification for the encryption of electronic data. It has been adopted by the U.S. government and is now used worldwide. AES is a symmetric-key algorithm, meaning the same key is used for both encrypting and decrypting the data. This paper defines the method to enhance the block and key length of the conventional AES.
This document discusses different types of balanced binary search trees including AVL trees, B-trees, and B+-trees. It provides examples of how to construct and perform operations on each type of tree. AVL trees balance after each insertion or deletion by performing tree rotations if needed. B-trees allow nodes to have more than two children and remain balanced by splitting full nodes. B+-trees are like B-trees but also allow sequential record traversal through leaf nodes. Threaded binary trees add threads to leaf nodes for efficient in-order traversal.
The document discusses binary search trees and their properties. It explains that a binary search tree is a binary tree where every node's left subtree contains values less than the node's value and the right subtree contains greater values. Operations like search, insert, delete can be done in O(h) time where h is the height of the tree. The height is O(log n) for balanced trees but can be O(n) for unbalanced trees. The document also provides examples of using a binary search tree to sort a set of numbers in O(n log n) time by building the BST and doing an inorder traversal.
Application of Matrices in real life | Matrices application | The MatricesSahilJhajharia
Matrices can be used to transform vectors by changing their magnitude and direction. Matrices are useful for applications like rotating vectors, solving systems of linear equations, and encoding/decoding messages for cryptography. As an example, a message can be encoded using a matrix, transmitted as encoded values, and then decoded by the receiver using the inverse matrix. Eigenvectors are vectors that only change in magnitude and not direction when multiplied by a matrix. They can be used to model real-world systems changing over time, like populations of humans and zombies.
Presentation given at DMZ about Data Structure Graphs.
Also known as Applying Social Network Analysis Techniques to Data Modeling and Data Architecture
OVERVIEW:
Introduction
Definition
Example of Threaded BT.
Types & Structure
One-way .
Double-way.
Structure.
Traversal
Algorithm for Traversal
Traversal Example
Inserting
Algorithm for Inserting
Inserting Example
Comparison With Binary Tree
Advantages and Disadvantages
Why Threaded BT are used?
Conclusion
Reference
The document discusses various data addressing modes used in microprocessors, including register, immediate, direct, indirect, base-plus-index, register relative, base relative-plus-index, and scaled-index addressing. It also covers program addressing modes like direct, relative, and indirect jumping and calling. Finally, it discusses stack addressing and how the push and pop instructions are used to place data onto and remove data from the stack.
Cryptography an application of vectors and matricesdianasc04
This document discusses cryptography and various encryption methods using matrices. It introduces shift ciphers, stretch ciphers, combination ciphers, and the Vigenere cipher which uses a keyword to shift between cipher alphabets. It provides examples of encoding and decoding messages with these ciphers and discusses how matrices can represent and manipulate encrypted data. It also considers the benefits and limitations of different encryption methods and how cryptography applies to fields like warfare.
This document provides an introduction and overview of a discrete mathematics course. It explains that discrete mathematics lays the mathematical foundations for many computer science topics. It lists common topics covered in the course like logic, sets, functions, counting, recursion, graphs and trees. It emphasizes that discrete mathematics deals with discrete, distinct objects like those used in computing. Mastering this subject involves actively practicing the concepts rather than just hearing or reading about them.
1) The document is an assignment submission on data structures covering queues, linked lists, and trees.
2) It discusses the concepts of queues including FIFO operations, applications such as call centers, and implementations using arrays or linked lists.
3) Linked lists are described as dynamic structures that connect nodes through pointers and are used for stacks, queues, and graphs. Common types include singly, doubly, and circular linked lists.
4) Trees are hierarchical structures where each node has a maximum of two children. Binary trees, their terminology, types such as full and balanced binary trees, and advantages are outlined.
O'Donnell Consulting Engineers evaluated vibrations in the framework supporting the Hubble Space Telescope. Tests and analyses showed the vibrations were caused by shortening of transverse bracing members. The company added lateral restraints to the bracing, easily solving the vibration problems. O'Donnell provides engineering design, analysis, materials consulting, and problem solving services to clients worldwide, with experience spanning industries and challenging issues like fabrication, fatigue, corrosion, and failure analysis. For over 40 years, the company has taken on complex technical problems for clients in various sectors.
The document summarizes the design process of Team RAPTOR for the 2015 NASA Robotic Mining Competition. It describes the systems engineering approach used, including defining requirements, developing a concept of operations, and conducting various design reviews. Key aspects of the preliminary and final robot design are discussed, such as using magnesium alloys for the structure to reduce weight and implementing a conveyor system for regolith transport to minimize tipping risks. The team's goal was to develop an autonomous robot called DAIR capable of deep icy regolith excavation.
Jacquelin L. Jenkins has over 30 years of experience in healthcare, including experience as a nurse, quality care coordinator, and community health educator. She has a proven track record of developing training protocols, leading teams, and building relationships across cultures. Jenkins holds a Master's degree in Organization Leadership and Bachelor's degrees in Health Science and Nursing. She is licensed as a registered nurse in California.
Sameer Dnyaneshwar is a 29-year-old magician and mind reader from India with over 22 years of experience. He has had a career in magic from a young age, following in his father's footsteps. Sameer performs various magic shows for corporate events and has been a regular performer on Carnival cruise lines in the US since 2010, entertaining international audiences. His mind reading show, called "Mentalini", is a popular 45-60 minute act that uses psychic abilities, telekinesis, and comedy to engage audiences.
Swapnil Madhukar Metkar's CV summarizes his professional experience and qualifications. He has over 3.8 years of experience working as a Customer Support Engineer for Sysnet Global Technology providing on-site support for State Bank of India. He holds qualifications including a JCHNP from Jetking, a BCom from YCMOU, and has skills in computer hardware, software installation, LAN setup, printer repair, and Linux server configuration. His objective is to obtain a position utilizing his interpersonal skills and experience to become a successful IT professional.
Dokumen tersebut memberikan informasi tentang Himpunan Mahasiswa Program Studi Pendidikan Guru Madrasah Ibtidaiyah (HMPS PGMI) di Sekolah Tinggi Agama Islam Negeri (STAIN) Pekalongan. Terdapat penjelasan singkat tentang visi, misi, struktur organisasi, dan program kerja HMPS PGMI STAIN Pekalongan.
Este documento define un proyecto como una idea para resolver un problema, explica que la gerencia es dirigir y administrar una empresa, y que los proyectos facilitan el proceso de toma de decisiones para inversiones al verificar la viabilidad técnica, comercial, económica, legal y financiera de una iniciativa. Finalmente, enumera diferentes tipos de proyectos como simples, complejos, públicos, privados y mixtos.
The document summarizes the professional experience and qualifications of Arya Narayan. She is seeking a responsible career opportunity utilizing her engineering, marketing, administration, and management skills. She has over 2 years of experience in service coordination and business administration. Her previous roles include Marketing Support Executive at The New Indian Express, where she sold advertising space, managed campaigns, and identified new clients. She also worked as a Business Development Specialist, developing new business by identifying prospects and needs. She has an MBA and engineering degrees and is proficient in MS Office, CRM software, and SQL.
Enjoy your sleep time with this printed night wear gown for women from Indiatrendzs. Made from Cotton, this nighty or night dress as we may call it is easy on the skin and features short sleeve. Look good and feel relaxed by wearing this Printed Sleepwear.
El estudio de caso es una técnica de aprendizaje en la que los estudiantes analizan una situación problemática de la vida real en pequeños grupos. Los estudiantes deben comprender y resolver el problema a través de la discusión. El estudio de caso desarrolla el espíritu crítico de los estudiantes y los prepara para tomar decisiones mediante la defensa de sus argumentos.
Scotland's capital is Edinburgh. Scottish whiskies are renowned worldwide, and traditional dishes include broth Scoth. Queen Elizabeth II is the Queen of Scotland. Christianity is the official religion, with the largest denominations being Presbyterian, Roman Catholic, and other Christian faiths. Bagpipe music is deeply rooted in Scottish Gaelic culture and divided into styles for large gatherings and small performances. The national motto is "Nemo me impune lacessit", which means "No one provokes me with impunity". The Loch Ness Monster is a legendary creature that some believe inhabits Loch Ness in the Scottish Highlands. Kilts are the traditional dress of Scottish men and boys, typically made of tartan wool and worn
The document provides a table of contents and questions on various technical aptitude topics such as data structures, C, C++, quantitative aptitude, UNIX concepts, RDBMS concepts, SQL, computer networks, and operating systems. It contains 35 questions related to data structures concepts like linked lists, trees, graphs, hashing etc. and their applications.
This document provides an overview of various data structures concepts and their applications. It discusses topics like data structures used in different areas like RDBMS, networks, and hierarchical models. Common data structures include arrays, linked lists, stacks, queues, trees and graphs. It also summarizes sorting and searching algorithms like quicksort, different tree traversals, and hashing techniques for resolving collisions. The key data structure used for storage in RDBMS is the B+ tree due to its efficient search capability.
This document provides an overview of topics related to data structures aptitude, including:
- Common data structures used in areas like databases, networks, and operating systems.
- Examples of data structures questions, such as converting expressions to prefix and postfix notation.
- Algorithms related to data structures, including sorting, tree traversal, and hashing.
- Detailed examples are provided, such as steps to implement quicksort on an array of numbers.
This document provides a table of contents and overview of topics related to technical aptitude questions, including data structures, C/C++ programming, quantitative aptitude, UNIX concepts, relational database management systems (RDBMS), SQL, computer networks, and operating systems. It discusses key data structure concepts like data structures used in different areas, pointers for heterogeneous linked lists, priority queues, recursion, sorting methods, trees, and hashing functions. It also includes examples of various data structure problems and their solutions.
A graph is a non-linear data structure consisting of nodes and edges where the nodes are connected via edges. There are different ways to represent graphs including using an adjacency matrix or adjacency lists. Common graph terminology includes vertices, edges, degree, and traversal algorithms like depth-first search (DFS) and breadth-first search (BFS) which are used to search graphs. DFS uses a stack and explores nodes as deep as possible before backtracking while BFS uses a queue and explores all neighbor nodes at the present depth before moving deeper.
This is the second lecture in the CS 6212 class. Covers asymptotic notation and data structures. Also outlines the coming lectures wherein we will study the various algorithm design techniques.
This document defines and provides examples of trees and binary trees. It begins by defining trees as hierarchical data structures with nodes and edges. It then provides definitions for terms like path, forest, ordered tree, height, and multiway tree. It specifically defines binary trees as having two children per node. The document gives examples and properties of binary trees, including full, complete, and binary search trees. It also explains linear and linked representations of binary trees and different traversal methods like preorder, postorder and inorder. Finally, it provides examples of insertion and deletion operations in binary search trees.
This document provides an overview of trees and graphs as data structures. It discusses binary trees, including their implementation and operations like searching. Binary search trees are described, with examples of insertion and searching. Graphs are then introduced, including undirected and directed graphs. Graph representations in Python are shown. Examples of an undirected weighted graph, minimal spanning tree, and shortest paths are provided. Finally, it notes some common graph algorithms.
The document discusses binary tree representation and traversal methods. It provides two main representations of binary trees - array and linked representations. It also describes different types of binary trees like full, complete, left-skewed, and right-skewed trees. The document then explains three common traversal techniques for binary trees - inorder, preorder, and postorder traversals. Algorithms and code snippets are given for each traversal. Finally, applications of binary tree traversals are discussed along with expression trees and converting infix to postfix notation using a stack.
A simple and fast exact clustering algorithm defined for complexAlexander Decker
This document proposes a new clustering method for complex networks based on prime numbers. The method defines clusters of nodes that have the same number of input/output connections and paths of equal length. It encodes the complete paths of each node with prime numbers to calculate a unique CPS-code for that node. Nodes with the same CPS-code are grouped into the same cluster. The algorithm was tested on networks with up to 500 nodes and showed fast performance, analyzing networks in seconds or minutes. The simple method allows classification of nodes in complex networks for applications across different scientific fields.
A simple and fast exact clustering algorithm defined for complexAlexander Decker
This document proposes a new clustering method for complex networks based on prime numbers. It defines nodes clusters as groups of nodes with the same complete path sets (CPS), where a complete path is a path from a node that includes each other node only once. Each complete path is assigned a prime number as a code based on its length. A node's CPS-code is the product of the codes for all its complete paths. Nodes are clustered together if they have the same or similar CPS-codes. The method was tested on a network with 500 nodes and over 100,000 connections, which was clustered in 80 seconds.
The binary search is faster than the sequential search. The complexity of binary search is O(log n) whereas the complexity of a sequential search is O(n). Stacks are used to evaluate algebraic or arithmetic expressions using prefix or postfix notations. Heap sort involves creating a max heap from the array and then replacing the root with the last element and rebuilding the heap for the remaining elements, repeating this process to sort the entire array.
The document discusses data structures and linked lists. It defines data structures as logical ways of organizing and storing data in computer memory for efficient use. Linked lists are introduced as a linear data structure where elements are connected via pointers and can grow/shrink dynamically. Algorithms for traversing, inserting, and deleting nodes from singly linked lists using both recursion and iteration are provided with pseudocode.
1. The document discusses data structures and provides 23 questions and answers about data structures.
2. It covers topics like common data structures, their applications in areas like databases and operating systems, implementations of linked lists and queues, sorting algorithms, tree and graph traversals, hashing techniques and more.
3. The questions aim to test the reader's understanding of fundamental data structures concepts.
The document provides information about data structures and algorithms. It contains questions and answers on topics such as data structures, trees, graphs, hashing, sorting, and file structures. Examples of code, expressions, and graphs are included to illustrate different data structures and algorithms.
The document provides information about data structures, including definitions of key terms, examples of different data structure types, and operations that can be performed on data structures.
It begins by defining a data structure as a collection of elements and operations on those elements. Linear data structures like stacks, queues, and linked lists are described, where elements are arranged sequentially. Non-linear structures like trees and graphs are also mentioned.
Common operations on data structures include creation, insertion, deletion, searching, sorting, and reversing elements. Abstract data types are defined, and several applications of data structures in areas like operating systems, databases, and artificial intelligence are listed. Specific data structure types like linked lists, stacks, and queues are then defined
This document discusses graphs and their representation in code. It defines graphs as consisting of vertices and edges, with edges specified as pairs of vertices. It distinguishes between directed and undirected graphs. Key graph terms like paths, cycles, and connectivity are defined. Real-world systems that can be modeled as graphs are given as an example. The document then discusses representing vertices and edges in code, choosing an adjacency matrix to represent the edges in the graph.
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for sbi so Ds c c++ unix rdbms sql cn os
1. Data Structures Aptitude
1. What is data structure?
A data structure is a way of organizing data that considers not only the items
stored, but also their relationship to each other. Advance knowledge about the
relationship between data items allows designing of efficient algorithms for the
manipulation of data.
2. List out the areas in which data structures are applied extensively?
Compiler Design,
Operating System,
Database Management System,
Statistical analysis package,
Numerical Analysis,
Graphics,
Artificial Intelligence,
Simulation
3. What are the major data structures used in the following areas : RDBMS, Network
data model & Hierarchical data model.
RDBMS – Array (i.e. Array of structures)
Network data model – Graph
Hierarchical data model – Trees
4. If you are using C language to implement the heterogeneous linked list, what pointer
type will you use?
The heterogeneous linked list contains different data types in its nodes and we
need a link, pointer to connect them. It is not possible to use ordinary pointers for this. So
we go for void pointer. Void pointer is capable of storing pointer to any type as it is a
generic pointer type.
5. Minimum number of queues needed to implement the priority queue?
Two. One queue is used for actual storing of data and another for storing
priorities.
6. What is the data structures used to perform recursion?
Data Structures Aptitude
2. Stack. Because of its LIFO (Last In First Out) property it remembers its ‘caller’ so
knows whom to return when the function has to return. Recursion makes use of system
stack for storing the return addresses of the function calls.
Every recursive function has its equivalent iterative (non-recursive) function.
Even when such equivalent iterative procedures are written, explicit stack is to be used.
7. What are the notations used in Evaluation of Arithmetic Expressions using prefix and
postfix forms?
Polish and Reverse Polish notations.
8. Convert the expression ((A + B) * C – (D – E) ^ (F + G)) to equivalent Prefix and
Postfix notations.
Prefix Notation:
^ - * +ABC - DE + FG
Postfix Notation:
AB + C * DE - - FG + ^
9. Sorting is not possible by using which of the following methods?
(a) Insertion
(b) Selection
(c) Exchange
(d) Deletion
(d) Deletion.
Using insertion we can perform insertion sort, using selection we can perform
selection sort, using exchange we can perform the bubble sort (and other similar sorting
methods). But no sorting method can be done just using deletion.
10. A binary tree with 20 nodes has null branches?
21
Let us take a tree with 5 nodes (n=5)
It will have only 6 (ie,5+1) null branches. In general,
A binary tree with n nodes has exactly n+1 null nodes.
11. What are the methods available in storing sequential files ?
Straight merging,
2
Null Branches
3. Natural merging,
Polyphase sort,
Distribution of Initial runs.
12. How many different trees are possible with 10 nodes ?
1014
For example, consider a tree with 3 nodes(n=3), it will have the maximum
combination of 5 different (ie, 23
- 3 = 5) trees.
i ii iii iv v
In general:
If there are n nodes, there exist 2n
-n different trees.
13. List out few of the Application of tree data-structure?
The manipulation of Arithmetic expression,
Symbol Table construction,
Syntax analysis.
14. List out few of the applications that make use of Multilinked Structures?
Sparse matrix,
Index generation.
15. In tree construction which is the suitable efficient data structure?
(a) Array (b) Linked list (c) Stack (d) Queue (e) none
(b) Linked list
16. What is the type of the algorithm used in solving the 8 Queens problem?
Backtracking
17. In an AVL tree, at what condition the balancing is to be done?
If the ‘pivotal value’ (or the ‘Height factor’) is greater than 1 or less than –1.
18. What is the bucket size, when the overlapping and collision occur at same time?
3
4. One. If there is only one entry possible in the bucket, when the collision occurs,
there is no way to accommodate the colliding value. This results in the overlapping of
values.
19. Traverse the given tree using Inorder, Preorder and Postorder traversals.
Inorder : D H B E A F C I G J
Preorder: A B D H E C F G I J
Postorder: H D E B F I J G C A
20. There are 8, 15, 13, 14 nodes were there in 4 different trees. Which of them could
have formed a full binary tree?
15.
In general:
There are 2n
-1 nodes in a full binary tree.
By the method of elimination:
Full binary trees contain odd number of nodes. So there cannot be full
binary trees with 8 or 14 nodes, so rejected. With 13 nodes you can form a complete
binary tree but not a full binary tree. So the correct answer is 15.
Note:
Full and Complete binary trees are different. All full binary trees are complete
binary trees but not vice versa.
21. In the given binary tree, using array you can store the node 4 at which location?
4
A
B C
D E F G
H I J
Given tree:
1
2 3
4
5
5. At location 6
1 2 3 - - 4 - - 5
Root LC1 RC1 LC2 RC2 LC3 RC3 LC4 RC4
where LCn means Left Child of node n and RCn means Right Child of node n
22. Sort the given values using Quick Sort?
65 70 75 80 85 60 55 50 45
Sorting takes place from the pivot value, which is the first value of the given
elements, this is marked bold. The values at the left pointer and right pointer are indicated
using L
and R
respectively.
65 70L
75 80 85 60 55 50 45R
Since pivot is not yet changed the same process is continued after interchanging the
values at L
and R
positions
65 45 75L
80 85 60 55 50R
70
65 45 50 80L
85 60 55R
75 70
65 45 50 55 85L
60R
80 75 70
65 45 50 55 60R
85L
80 75 70
When the L and R pointers cross each other the pivot value is interchanged with the value
at right pointer. If the pivot is changed it means that the pivot has occupied its original
position in the sorted order (shown in bold italics) and hence two different arrays are
formed, one from start of the original array to the pivot position-1 and the other from
pivot position+1 to end.
60L
45 50 55R
65 85L
80 75 70R
5
6. 55L
45 50R
60 65 70R
80L
75 85
50L
45R
55 60 65 70 80L
75R
85
In the next pass we get the sorted form of the array.
45 50 55 60 65 70 75 80 85
23. For the given graph, draw the DFS and BFS?
BFS: A X G H P E M Y J
DFS: A X H P E Y M J G
24. Classify the Hashing Functions based on the various methods by which the key value
is found.
Direct method,
Subtraction method,
Modulo-Division method,
Digit-Extraction method,
Mid-Square method,
Folding method,
Pseudo-random method.
25. What are the types of Collision Resolution Techniques and the methods used in each
of the type?
Open addressing (closed hashing),
The methods used include:
Overflow block,
Closed addressing (open hashing)
The methods used include:
Linked list,
6
A
HX
G P
E
Y
M J
The given graph:
7. Binary tree…
26. In RDBMS, what is the efficient data structure used in the internal storage
representation?
B+ tree. Because in B+ tree, all the data is stored only in leaf nodes, that makes
searching easier. This corresponds to the records that shall be stored in leaf nodes.
27. Draw the B-tree of order 3 created by inserting the following data arriving in
sequence – 92 24 6 7 11 8 22 4 5 16 19 20 78
28.Of the following tree structure, which is, efficient considering space and
time complexities?
(a) Incomplete Binary Tree
(b) Complete Binary Tree
(c) Full Binary Tree
(b) Complete Binary Tree.
By the method of elimination:
Full binary tree loses its nature when operations of insertions and deletions
are done. For incomplete binary trees, extra storage is required and overhead of NULL
node checking takes place. So complete binary tree is the better one since the property of
complete binary tree is maintained even after operations like additions and deletions are
done on it.
29. What is a spanning Tree?
A spanning tree is a tree associated with a network. All the nodes of the graph
appear on the tree once. A minimum spanning tree is a spanning tree organized so that
the total edge weight between nodes is minimized.
30. Does the minimum spanning tree of a graph give the shortest distance between any 2
specified nodes?
No.
Minimal spanning tree assures that the total weight of the tree is kept at its
minimum. But it doesn’t mean that the distance between any two nodes involved in the
minimum-spanning tree is minimum.
7
11 -
5 7 19 24
4 - 6 - 8 - 16 - 20 22 78 92
8. 31. Convert the given graph with weighted edges to minimal spanning tree.
the equivalent minimal spanning tree is:
32. Which is the simplest file structure?
(a) Sequential
(b) Indexed
(c) Random
(a) Sequential
33. Whether Linked List is linear or Non-linear data structure?
According to Access strategies Linked list is a linear one.
According to Storage Linked List is a Non-linear one.
34. Draw a binary Tree for the expression :
A * B - (C + D) * (P / Q)
8
1 3
2 4
5410
600
200
400
310
1421
2985
612
-
* *
A B + /
C PD Q
1
2
3
4 5
410 612
200
310
9. 35. For the following COBOL code, draw the Binary tree?
01 STUDENT_REC.
02 NAME.
03 FIRST_NAME PIC X(10).
03 LAST_NAME PIC X(10).
02 YEAR_OF_STUDY.
03 FIRST_SEM PIC XX.
03 SECOND_SEM PIC XX.
9
STUDENT_REC
NAME YEAR_OF_STUDY
FIRST_NAME LAST_NAME FIRST_SEM SECOND_SEM
01
02 02
03 03 03 03
10. C Aptitude
Note : All the programs are tested under Turbo C/C++ compilers.
It is assumed that,
Programs run under DOS environment,
The underlying machine is an x86 system,
Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions
(for example sizeof(int) == 2 may be assumed).
Predict the output or error(s) for the following:
1. void main()
{
int const * p=5;
printf("%d",++(*p));
}
Answer:
Compiler error: Cannot modify a constant value.
Explanation:
p is a pointer to a "constant integer". But we tried to change the value of
the "constant integer".
2. main()
{
char s[ ]="man";
int i;
for(i=0;s[ i ];i++)
printf("n%c%c%c%c",s[ i ],*(s+i),*(i+s),i[s]);
}
Answer:
mmmm
aaaa
nnnn
Explanation:
s[i], *(i+s), *(s+i), i[s] are all different ways of expressing the same idea.
Generally array name is the base address for that array. Here s is the base address. i is the
10
C Aptitude
11. index number/displacement from the base address. So, indirecting it with * is same as
s[i]. i[s] may be surprising. But in the case of C it is same as s[i].
3. main()
{
float me = 1.1;
double you = 1.1;
if(me==you)
printf("I love U");
else
printf("I hate U");
}
Answer:
I hate U
Explanation:
For floating point numbers (float, double, long double) the values cannot
be predicted exactly. Depending on the number of bytes, the precession with of the value
represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9
with less precision than long double.
Rule of Thumb:
Never compare or at-least be cautious when using floating point numbers
with relational operators (== , >, <, <=, >=,!= ) .
4. main()
{
static int var = 5;
printf("%d ",var--);
if(var)
main();
}
Answer:
5 4 3 2 1
Explanation:
When static storage class is given, it is initialized once. The change in the
value of a static variable is retained even between the function calls. Main is also treated
like any other ordinary function, which can be called recursively.
5. main()
{
int c[ ]={2.8,3.4,4,6.7,5};
int j,*p=c,*q=c;
for(j=0;j<5;j++) {
printf(" %d ",*c);
++q; }
for(j=0;j<5;j++){
printf(" %d ",*p);
++p; }
11
12. }
Answer:
2 2 2 2 2 2 3 4 6 5
Explanation:
Initially pointer c is assigned to both p and q. In the first loop, since only
q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is
incremented. So the values 2 3 4 6 5 will be printed.
6. main()
{
extern int i;
i=20;
printf("%d",i);
}
Answer:
Linker Error : Undefined symbol '_i'
Explanation:
extern storage class in the following declaration,
extern int i;
specifies to the compiler that the memory for i is allocated in some other program and
that address will be given to the current program at the time of linking. But linker finds
that no other variable of name i is available in any other program with memory space
allocated for it. Hence a linker error has occurred .
7. main()
{
int i=-1,j=-1,k=0,l=2,m;
m=i++&&j++&&k++||l++;
printf("%d %d %d %d %d",i,j,k,l,m);
}
Answer:
0 0 1 3 1
Explanation :
Logical operations always give a result of 1 or 0 . And also the logical
AND (&&) operator has higher priority over the logical OR (||) operator. So the
expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0
(-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR
operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value
of m is 1. The values of other variables are also incremented by 1.
8. main()
{
char *p;
printf("%d %d ",sizeof(*p),sizeof(p));
12
13. }
Answer:
1 2
Explanation:
The sizeof() operator gives the number of bytes taken by its operand. P is
a character pointer, which needs one byte for storing its value (a character). Hence
sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character
pointer sizeof(p) gives 2.
9. main()
{
int i=3;
switch(i)
{
default:printf("zero");
case 1: printf("one");
break;
case 2:printf("two");
break;
case 3: printf("three");
break;
}
}
Answer :
three
Explanation :
The default case can be placed anywhere inside the loop. It is executed
only when all other cases doesn't match.
10. main()
{
printf("%x",-1<<4);
}
Answer:
fff0
Explanation :
-1 is internally represented as all 1's. When left shifted four times the least
significant 4 bits are filled with 0's.The %x format specifier specifies that the integer
value be printed as a hexadecimal value.
11. main()
{
char string[]="Hello World";
display(string);
}
13
14. void display(char *string)
{
printf("%s",string);
}
Answer:
Compiler Error : Type mismatch in redeclaration of function display
Explanation :
In third line, when the function display is encountered, the compiler
doesn't know anything about the function display. It assumes the arguments and return
types to be integers, (which is the default type). When it sees the actual function display,
the arguments and type contradicts with what it has assumed previously. Hence a compile
time error occurs.
12. main()
{
int c=- -2;
printf("c=%d",c);
}
Answer:
c=2;
Explanation:
Here unary minus (or negation) operator is used twice. Same maths rules
applies, ie. minus * minus= plus.
Note:
However you cannot give like --2. Because -- operator can only be
applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.
13. #define int char
main()
{
int i=65;
printf("sizeof(i)=%d",sizeof(i));
}
Answer:
sizeof(i)=1
Explanation:
Since the #define replaces the string int by the macro char
14. main()
{
int i=10;
i=!i>14;
Printf ("i=%d",i);
}
Answer:
i=0
14
15. Explanation:
In the expression !i>14 , NOT (!) operator has more precedence than ‘ >’
symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false
(zero).
15. #include<stdio.h>
main()
{
char s[]={'a','b','c','n','c','0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
77
Explanation:
p is pointing to character 'n'. str1 is pointing to character 'a' ++*p. "p is pointing
to 'n' and that is incremented by one." the ASCII value of 'n' is 10, which is then
incremented to 11. The value of ++*p is 11. ++*str1, str1 is pointing to 'a' that is
incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98.
Now performing (11 + 98 – 32), we get 77("M");
So we get the output 77 :: "M" (Ascii is 77).
16. #include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d----%d",*p,*q);
}
Answer:
SomeGarbageValue---1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays, but you are trying to
access the third 2D(which you are not declared) it will print garbage values. *q=***a
starting address of a is assigned integer pointer. Now q is pointing to starting address of a.
If you print *q, it will print first element of 3D array.
17. #include<stdio.h>
main()
15
16. {
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
You should not initialize variables in declaration
18. #include<stdio.h>
main()
{
struct xx
{
int x;
struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
The structure yy is nested within structure xx. Hence, the elements are of
yy are to be accessed through the instance of structure xx, which needs an instance of yy
to be known. If the instance is created after defining the structure the compiler will not
know about the instance relative to xx. Hence for nested structure yy you have to declare
member.
19. main()
{
printf("nab");
printf("bsi");
printf("rha");
}
Answer:
hai
16
17. Explanation:
n - newline
b - backspace
r - linefeed
20. main()
{
int i=5;
printf("%d%d%d%d%d%d",i++,i--,++i,--i,i);
}
Answer:
45545
Explanation:
The arguments in a function call are pushed into the stack from left to
right. The evaluation is by popping out from the stack. and the evaluation is from right to
left, hence the result.
21. #define square(x) x*x
main()
{
int i;
i = 64/square(4);
printf("%d",i);
}
Answer:
64
Explanation:
the macro call square(4) will substituted by 4*4 so the expression becomes
i = 64/4*4 . Since / and * has equal priority the expression will be evaluated as (64/4)*4
i.e. 16*4 = 64
22. main()
{
char *p="hai friends",*p1;
p1=p;
while(*p!='0') ++*p++;
printf("%s %s",p,p1);
}
Answer:
ibj!gsjfoet
Explanation:
++*p++ will be parse in the given order
*p that is value at the location currently pointed by p will be taken
++*p the retrieved value will be incremented
when ; is encountered the location will be incremented that is p++ will be executed
17
18. Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by
executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so
on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!
gsjfoet” and since p reaches ‘0’ and p1 points to p thus p1doesnot print anything.
23. #include <stdio.h>
#define a 10
main()
{
#define a 50
printf("%d",a);
}
Answer:
50
Explanation:
The preprocessor directives can be redefined anywhere in the program. So
the most recently assigned value will be taken.
24. #define clrscr() 100
main()
{
clrscr();
printf("%dn",clrscr());
}
Answer:
100
Explanation:
Preprocessor executes as a seperate pass before the execution of the
compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler
looks like this :
main()
{
100;
printf("%dn",100);
}
Note:
100; is an executable statement but with no action. So it doesn't give any
problem
25. main()
{
printf("%p",main);
}
Answer:
Some address will be printed.
Explanation:
18
19. Function names are just addresses (just like array names are addresses).
main() is also a function. So the address of function main will be printed. %p in printf
specifies that the argument is an address. They are printed as hexadecimal numbers.
27) main()
{
clrscr();
}
clrscr();
Answer:
No output/error
Explanation:
The first clrscr() occurs inside a function. So it becomes a function call. In
the second clrscr(); is a function declaration (because it is not inside any
function).
28) enum colors {BLACK,BLUE,GREEN}
main()
{
printf("%d..%d..%d",BLACK,BLUE,GREEN);
return(1);
}
Answer:
0..1..2
Explanation:
enum assigns numbers starting from 0, if not explicitly defined.
29) void main()
{
char far *farther,*farthest;
printf("%d..%d",sizeof(farther),sizeof(farthest));
}
Answer:
4..2
Explanation:
the second pointer is of char type and not a far pointer
30) main()
{
int i=400,j=300;
printf("%d..%d");
19
20. }
Answer:
400..300
Explanation:
printf takes the values of the first two assignments of the program. Any
number of printf's may be given. All of them take only the first two
values. If more number of assignments given in the program,then printf
will take garbage values.
31) main()
{
char *p;
p="Hello";
printf("%cn",*&*p);
}
Answer:
H
Explanation:
* is a dereference operator & is a reference operator. They can be
applied any number of times provided it is meaningful. Here p points to
the first character in the string "Hello". *p dereferences it and so its value
is H. Again & references it to an address and * dereferences it to the value
H.
32) main()
{
int i=1;
while (i<=5)
{
printf("%d",i);
if (i>2)
goto here;
i++;
}
}
fun()
{
here:
printf("PP");
}
Answer:
Compiler error: Undefined label 'here' in function main
Explanation:
Labels have functions scope, in other words The scope of the labels is
limited to functions . The label 'here' is available in function fun() Hence it
is not visible in function main.
20
21. 33) main()
{
static char names[5][20]={"pascal","ada","cobol","fortran","perl"};
int i;
char *t;
t=names[3];
names[3]=names[4];
names[4]=t;
for (i=0;i<=4;i++)
printf("%s",names[i]);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
Array names are pointer constants. So it cannot be modified.
34) void main()
{
int i=5;
printf("%d",i++ + ++i);
}
Answer:
Output Cannot be predicted exactly.
Explanation:
Side effects are involved in the evaluation of i
35) void main()
{
int i=5;
printf("%d",i+++++i);
}
Answer:
Compiler Error
Explanation:
The expression i+++++i is parsed as i ++ ++ + i which is an illegal
combination of operators.
36) #include<stdio.h>
main()
{
int i=1,j=2;
switch(i)
{
case 1: printf("GOOD");
break;
21
22. case j: printf("BAD");
break;
}
}
Answer:
Compiler Error: Constant expression required in function main.
Explanation:
The case statement can have only constant expressions (this implies that
we cannot use variable names directly so an error).
Note:
Enumerated types can be used in case statements.
37) main()
{
int i;
printf("%d",scanf("%d",&i)); // value 10 is given as input here
}
Answer:
1
Explanation:
Scanf returns number of items successfully read and not 1/0. Here 10 is
given as input which should have been scanned successfully. So number
of items read is 1.
38) #define f(g,g2) g##g2
main()
{
int var12=100;
printf("%d",f(var,12));
}
Answer:
100
39) main()
{
int i=0;
for(;i++;printf("%d",i)) ;
printf("%d",i);
}
Answer:
1
Explanation:
before entering into the for loop the checking condition is "evaluated".
Here it evaluates to 0 (false) and comes out of the loop, and i is
incremented (note the semicolon after the for loop).
22
23. 40) #include<stdio.h>
main()
{
char s[]={'a','b','c','n','c','0'};
char *p,*str,*str1;
p=&s[3];
str=p;
str1=s;
printf("%d",++*p + ++*str1-32);
}
Answer:
M
Explanation:
p is pointing to character 'n'.str1 is pointing to character 'a' ++*p
meAnswer:"p is pointing to 'n' and that is incremented by one." the ASCII
value of 'n' is 10. then it is incremented to 11. the value of ++*p is 11. +
+*str1 meAnswer:"str1 is pointing to 'a' that is incremented by 1 and it
becomes 'b'. ASCII value of 'b' is 98. both 11 and 98 is added and result is
subtracted from 32.
i.e. (11+98-32)=77("M");
41) #include<stdio.h>
main()
{
struct xx
{
int x=3;
char name[]="hello";
};
struct xx *s=malloc(sizeof(struct xx));
printf("%d",s->x);
printf("%s",s->name);
}
Answer:
Compiler Error
Explanation:
Initialization should not be done for structure members inside the structure
declaration
42) #include<stdio.h>
main()
{
struct xx
{
int x;
23
24. struct yy
{
char s;
struct xx *p;
};
struct yy *q;
};
}
Answer:
Compiler Error
Explanation:
in the end of nested structure yy a member have to be declared.
43) main()
{
extern int i;
i=20;
printf("%d",sizeof(i));
}
Answer:
Linker error: undefined symbol '_i'.
Explanation:
extern declaration specifies that the variable i is defined somewhere else.
The compiler passes the external variable to be resolved by the linker. So
compiler doesn't find an error. During linking the linker searches for the
definition of i. Since it is not found the linker flags an error.
44) main()
{
printf("%d", out);
}
int out=100;
Answer:
Compiler error: undefined symbol out in function main.
Explanation:
The rule is that a variable is available for use from the point of declaration.
Even though a is a global variable, it is not available for main. Hence an
error.
45) main()
{
extern out;
printf("%d", out);
}
int out=100;
Answer:
24
25. 100
Explanation:
This is the correct way of writing the previous program.
46) main()
{
show();
}
void show()
{
printf("I'm the greatest");
}
Answer:
Compier error: Type mismatch in redeclaration of show.
Explanation:
When the compiler sees the function show it doesn't know anything about
it. So the default return type (ie, int) is assumed. But when compiler sees
the actual definition of show mismatch occurs since it is declared as void.
Hence the error.
The solutions are as follows:
1. declare void show() in main() .
2. define show() before main().
3. declare extern void show() before the use of show().
47) main( )
{
int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
printf(“%u %u %u %d n”,a,*a,**a,***a);
printf(“%u %u %u %d n”,a+1,*a+1,**a+1,***a+1);
}
Answer:
100, 100, 100, 2
114, 104, 102, 3
Explanation:
The given array is a 3-D one. It can also be viewed as a 1-D array.
2 4 7 8 3 4 2 2 2 3 3 4
100 102 104 106 108 110 112 114 116 118 120 122
thus, for the first printf statement a, *a, **a give address of first element .
since the indirection ***a gives the value. Hence, the first line of the
output.
for the second printf a+1 increases in the third dimension thus points to
value at 114, *a+1 increments in second dimension thus points to 104, **a
25
26. +1 increments the first dimension thus points to 102 and ***a+1 first gets
the value at first location and then increments it by 1. Hence, the output.
48) main( )
{
int a[ ] = {10,20,30,40,50},j,*p;
for(j=0; j<5; j++)
{
printf(“%d” ,*a);
a++;
}
p = a;
for(j=0; j<5; j++)
{
printf(“%d ” ,*p);
p++;
}
}
Answer:
Compiler error: lvalue required.
Explanation:
Error is in line with statement a++. The operand must be an lvalue and
may be of any of scalar type for the any operator, array name only when
subscripted is an lvalue. Simply array name is a non-modifiable lvalue.
49) main( )
{
static int a[ ] = {0,1,2,3,4};
int *p[ ] = {a,a+1,a+2,a+3,a+4};
int **ptr = p;
ptr++;
printf(“n %d %d %d”, ptr-p, *ptr-a, **ptr);
*ptr++;
printf(“n %d %d %d”, ptr-p, *ptr-a, **ptr);
*++ptr;
printf(“n %d %d %d”, ptr-p, *ptr-a, **ptr);
++*ptr;
printf(“n %d %d %d”, ptr-p, *ptr-a, **ptr);
}
Answer:
111
222
333
344
Explanation:
26
27. Let us consider the array and the two pointers with some address
a
0 1 2 3 4
100 102 104 106 108
p
100 102 104 106 108
1000 1002 1004 1006 1008
ptr
1000
2000
After execution of the instruction ptr++ value in ptr becomes 1002, if
scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting
location of array p, (1002 – 1000) / (scaling factor) = 1, *ptr – a = value at
address pointed by ptr – starting value of array a, 1002 has a value 102 so
the value is (102 – 100)/(scaling factor) = 1, **ptr is the value stored in
the location pointed by the pointer of ptr = value pointed by value pointed
by 1002 = value pointed by 102 = 1. Hence the output of the firs printf is
1, 1, 1.
After execution of *ptr++ increments value of the value in ptr by scaling
factor, so it becomes1004. Hence, the outputs for the second printf are ptr
– p = 2, *ptr – a = 2, **ptr = 2.
After execution of *++ptr increments value of the value in ptr by scaling
factor, so it becomes1004. Hence, the outputs for the third printf are ptr –
p = 3, *ptr – a = 3, **ptr = 3.
After execution of ++*ptr value in ptr remains the same, the value pointed
by the value is incremented by the scaling factor. So the value in array p at
location 1006 changes from 106 10 108,. Hence, the outputs for the fourth
printf are ptr – p = 1006 – 1000 = 3, *ptr – a = 108 – 100 = 4, **ptr = 4.
50) main( )
{
char *q;
int j;
for (j=0; j<3; j++) scanf(“%s” ,(q+j));
for (j=0; j<3; j++) printf(“%c” ,*(q+j));
for (j=0; j<3; j++) printf(“%s” ,(q+j));
}
Explanation:
Here we have only one pointer to type char and since we take input in the
same pointer thus we keep writing over in the same location, each time
shifting the pointer value by 1. Suppose the inputs are MOUSE, TRACK
and VIRTUAL. Then for the first input suppose the pointer starts at
location 100 then the input one is stored as
M O U S E 0
When the second input is given the pointer is incremented as j value
becomes 1, so the input is filled in memory starting from 101.
27
28. M T R A C K 0
The third input starts filling from the location 102
M T V I R T U A L 0
This is the final value stored .
The first printf prints the values at the position q, q+1 and q+2 = M T V
The second printf prints three strings starting from locations q, q+1, q+2
i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.
51) main( )
{
void *vp;
char ch = ‘g’, *cp = “goofy”;
int j = 20;
vp = &ch;
printf(“%c”, *(char *)vp);
vp = &j;
printf(“%d”,*(int *)vp);
vp = cp;
printf(“%s”,(char *)vp + 3);
}
Answer:
g20fy
Explanation:
Since a void pointer is used it can be type casted to any other type pointer.
vp = &ch stores address of char ch and the next statement prints the value
stored in vp after type casting it to the proper data type pointer. the output
is ‘g’. Similarly the output from second printf is ‘20’. The third printf
statement type casts it to print the string from the 4th
value hence the
output is ‘fy’.
52) main ( )
{
static char *s[ ] = {“black”, “white”, “yellow”, “violet”};
char **ptr[ ] = {s+3, s+2, s+1, s}, ***p;
p = ptr;
**++p;
printf(“%s”,*--*++p + 3);
}
Answer:
ck
Explanation:
In this problem we have an array of char pointers pointing to start of 4
strings. Then we have ptr which is a pointer to a pointer of type char and a
variable p which is a pointer to a pointer to a pointer of type char. p hold
the initial value of ptr, i.e. p = s+3. The next statement increment value in
p by 1 , thus now value of p = s+2. In the printf statement the expression
28
29. is evaluated *++p causes gets value s+1 then the pre decrement is
executed and we get s+1 – 1 = s . the indirection operator now gets the
value from the array of s and adds 3 to the starting address. The string is
printed starting from this position. Thus, the output is ‘ck’.
53) main()
{
int i, n;
char *x = “girl”;
n = strlen(x);
*x = x[n];
for(i=0; i<n; ++i)
{
printf(“%sn”,x);
x++;
}
}
Answer:
(blank space)
irl
rl
l
Explanation:
Here a string (a pointer to char) is initialized with a value “girl”. The
strlen function returns the length of the string, thus n has a value 4. The
next statement assigns value at the nth location (‘0’) to the first location.
Now the string becomes “0irl” . Now the printf statement prints the string
after each iteration it increments it starting position. Loop starts from 0 to
4. The first time x[0] = ‘0’ hence it prints nothing and pointer value is
incremented. The second time it prints from x[1] i.e “irl” and the third
time it prints “rl” and the last time it prints “l” and the loop terminates.
54) int i,j;
for(i=0;i<=10;i++)
{
j+=5;
assert(i<5);
}
Answer:
Runtime error: Abnormal program termination.
assert failed (i<5), <file name>,<line number>
Explanation:
asserts are used during debugging to make sure that certain conditions are
satisfied. If assertion fails, the program will terminate reporting the same.
After debugging use,
#undef NDEBUG
29
30. and this will disable all the assertions from the source code. Assertion
is a good debugging tool to make use of.
55) main()
{
int i=-1;
+i;
printf("i = %d, +i = %d n",i,+i);
}
Answer:
i = -1, +i = -1
Explanation:
Unary + is the only dummy operator in C. Where-ever it comes you can
just ignore it just because it has no effect in the expressions (hence the
name dummy operator).
56) What are the files which are automatically opened when a C file is executed?
Answer:
stdin, stdout, stderr (standard input,standard output,standard error).
57) what will be the position of the file marker?
a: fseek(ptr,0,SEEK_SET);
b: fseek(ptr,0,SEEK_CUR);
Answer :
a: The SEEK_SET sets the file position marker to the starting of the file.
b: The SEEK_CUR sets the file position marker to the current position
of the file.
58) main()
{
char name[10],s[12];
scanf(" "%[^"]"",s);
}
How scanf will execute?
Answer:
First it checks for the leading white space and discards it.Then it matches
with a quotation mark and then it reads all character upto another
quotation mark.
59) What is the problem with the following code segment?
while ((fgets(receiving array,50,file_ptr)) != EOF)
;
Answer & Explanation:
fgets returns a pointer. So the correct end of file check is checking for !=
NULL.
30
31. 60) main()
{
main();
}
Answer:
Runtime error : Stack overflow.
Explanation:
main function calls itself again and again. Each time the function is called
its return address is stored in the call stack. Since there is no condition to
terminate the function call, the call stack overflows at runtime. So it
terminates the program and results in an error.
61) main()
{
char *cptr,c;
void *vptr,v;
c=10; v=0;
cptr=&c; vptr=&v;
printf("%c%v",c,v);
}
Answer:
Compiler error (at line number 4): size of v is Unknown.
Explanation:
You can create a variable of type void * but not of type void, since void is
an empty type. In the second line you are creating variable vptr of type
void * and v of type void hence an error.
62) main()
{
char *str1="abcd";
char str2[]="abcd";
printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}
Answer:
2 5 5
Explanation:
In first sizeof, str1 is a character pointer so it gives you the size of the
pointer variable. In second sizeof the name str2 indicates the name of the
array whose size is 5 (including the '0' termination character). The third
sizeof is similar to the second one.
63) main()
{
char not;
not=!2;
31
32. printf("%d",not);
}
Answer:
0
Explanation:
! is a logical operator. In C the value 0 is considered to be the boolean
value FALSE, and any non-zero value is considered to be the boolean
value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0)
so it prints 0.
64) #define FALSE -1
#define TRUE 1
#define NULL 0
main() {
if(NULL)
puts("NULL");
else if(FALSE)
puts("TRUE");
else
puts("FALSE");
}
Answer:
TRUE
Explanation:
The input program to the compiler after processing by the preprocessor is,
main(){
if(0)
puts("NULL");
else if(-1)
puts("TRUE");
else
puts("FALSE");
}
Preprocessor doesn't replace the values given inside the double quotes.
The check by if condition is boolean value false so it goes to else. In
second if -1 is boolean value true hence "TRUE" is printed.
65) main()
{
int k=1;
printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}
Answer:
1==1 is TRUE
Explanation:
32
33. When two strings are placed together (or separated by white-space) they
are concatenated (this is called as "stringization" operation). So the string
is as if it is given as "%d==1 is %s". The conditional operator( ?: )
evaluates to "TRUE".
66) main()
{
int y;
scanf("%d",&y); // input given is 2000
if( (y%4==0 && y%100 != 0) || y%100 == 0 )
printf("%d is a leap year");
else
printf("%d is not a leap year");
}
Answer:
2000 is a leap year
Explanation:
An ordinary program to check if leap year or not.
67) #define max 5
#define int arr1[max]
main()
{
typedef char arr2[max];
arr1 list={0,1,2,3,4};
arr2 name="name";
printf("%d %s",list[0],name);
}
Answer:
Compiler error (in the line arr1 list = {0,1,2,3,4})
Explanation:
arr2 is declared of type array of size 5 of characters. So it can be used to
declare the variable name of the type arr2. But it is not the case of arr1.
Hence an error.
Rule of Thumb:
#defines are used for textual replacement whereas typedefs are used for
declaring new types.
68) int i=10;
main()
{
extern int i;
{
int i=20;
{
const volatile unsigned i=30;
33
34. printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
Answer:
30,20,10
Explanation:
'{' introduces new block and thus new scope. In the innermost block i is
declared as,
const volatile unsigned
which is a valid declaration. i is assumed of type int. So printf prints 30. In
the next block, i has value 20 and so printf prints 20. In the outermost
block, i is declared as extern, so no storage space is allocated for it. After
compilation is over the linker resolves it to global variable i (since it is the
only variable visible there). So it prints i's value as 10.
69) main()
{
int *j;
{
int i=10;
j=&i;
}
printf("%d",*j);
}
Answer:
10
Explanation:
The variable i is a block level variable and the visibility is inside that
block only. But the lifetime of i is lifetime of the function so it lives upto
the exit of main function. Since the i is still allocated space, *j prints the
value stored in i since j points i.
70) main()
{
int i=-1;
-i;
printf("i = %d, -i = %d n",i,-i);
}
Answer:
i = -1, -i = 1
Explanation:
34
35. -i is executed and this execution doesn't affect the value of i. In printf first
you just print the value of i. After that the value of the expression -i = -(-1)
is printed.
71) #include<stdio.h>
main()
{
const int i=4;
float j;
j = ++i;
printf("%d %f", i,++j);
}
Answer:
Compiler error
Explanation:
i is a constant. you cannot change the value of constant
72) #include<stdio.h>
main()
{
int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
int *p,*q;
p=&a[2][2][2];
*q=***a;
printf("%d..%d",*p,*q);
}
Answer:
garbagevalue..1
Explanation:
p=&a[2][2][2] you declare only two 2D arrays. but you are trying to
access the third 2D(which you are not declared) it will print garbage
values. *q=***a starting address of a is assigned integer pointer. now q is
pointing to starting address of a.if you print *q meAnswer:it will print first
element of 3D array.
73) #include<stdio.h>
main()
{
register i=5;
char j[]= "hello";
printf("%s %d",j,i);
}
Answer:
hello 5
Explanation:
35
36. if you declare i as register compiler will treat it as ordinary integer and it
will take integer value. i value may be stored either in register or in
memory.
74) main()
{
int i=5,j=6,z;
printf("%d",i+++j);
}
Answer:
11
Explanation:
the expression i+++j is treated as (i++ + j)
76) struct aaa{
struct aaa *prev;
int i;
struct aaa *next;
};
main()
{
struct aaa abc,def,ghi,jkl;
int x=100;
abc.i=0;abc.prev=&jkl;
abc.next=&def;
def.i=1;def.prev=&abc;def.next=&ghi;
ghi.i=2;ghi.prev=&def;
ghi.next=&jkl;
jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
x=abc.next->next->prev->next->i;
printf("%d",x);
}
Answer:
2
Explanation:
above all statements form a double circular linked list;
abc.next->next->prev->next->i
this one points to "ghi" node the value of at particular node is 2.
77) struct point
{
int x;
int y;
};
struct point origin,*pp;
main()
36
37. {
pp=&origin;
printf("origin is(%d%d)n",(*pp).x,(*pp).y);
printf("origin is (%d%d)n",pp->x,pp->y);
}
Answer:
origin is(0,0)
origin is(0,0)
Explanation:
pp is a pointer to structure. we can access the elements of the structure
either with arrow mark or with indirection operator.
Note:
Since structure point is globally declared x & y are initialized as zeroes
78) main()
{
int i=_l_abc(10);
printf("%dn",--i);
}
int _l_abc(int i)
{
return(i++);
}
Answer:
9
Explanation:
return(i++) it will first return i and then increments. i.e. 10 will be
returned.
79) main()
{
char *p;
int *q;
long *r;
p=q=r=0;
p++;
q++;
r++;
printf("%p...%p...%p",p,q,r);
}
Answer:
0001...0002...0004
Explanation:
++ operator when applied to pointers increments address according to
their corresponding data-types.
37
38. 80) main()
{
char c=' ',x,convert(z);
getc(c);
if((c>='a') && (c<='z'))
x=convert(c);
printf("%c",x);
}
convert(z)
{
return z-32;
}
Answer:
Compiler error
Explanation:
declaration of convert and format of getc() are wrong.
81) main(int argc, char **argv)
{
printf("enter the character");
getchar();
sum(argv[1],argv[2]);
}
sum(num1,num2)
int num1,num2;
{
return num1+num2;
}
Answer:
Compiler error.
Explanation:
argv[1] & argv[2] are strings. They are passed to the function sum without
converting it to integer values.
82) # include <stdio.h>
int one_d[]={1,2,3};
main()
{
int *ptr;
ptr=one_d;
ptr+=3;
printf("%d",*ptr);
}
Answer:
garbage value
38
39. Explanation:
ptr pointer is pointing to out of the array range of one_d.
83) # include<stdio.h>
aaa() {
printf("hi");
}
bbb(){
printf("hello");
}
ccc(){
printf("bye");
}
main()
{
int (*ptr[3])();
ptr[0]=aaa;
ptr[1]=bbb;
ptr[2]=ccc;
ptr[2]();
}
Answer:
bye
Explanation:
ptr is array of pointers to functions of return type int.ptr[0] is assigned to
address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc
respectively. ptr[2]() is in effect of writing ccc(), since ptr[2] points to ccc.
85) #include<stdio.h>
main()
{
FILE *ptr;
char i;
ptr=fopen("zzz.c","r");
while((i=fgetch(ptr))!=EOF)
printf("%c",i);
}
Answer:
contents of zzz.c followed by an infinite loop
Explanation:
The condition is checked against EOF, it should be checked against
NULL.
86) main()
{
int i =0;j=0;
39
40. if(i && j++)
printf("%d..%d",i++,j);
printf("%d..%d,i,j);
}
Answer:
0..0
Explanation:
The value of i is 0. Since this information is enough to determine the truth
value of the boolean expression. So the statement following the if
statement is not executed. The values of i and j remain unchanged and get
printed.
87) main()
{
int i;
i = abc();
printf("%d",i);
}
abc()
{
_AX = 1000;
}
Answer:
1000
Explanation:
Normally the return value from the function is through the information
from the accumulator. Here _AH is the pseudo global variable denoting
the accumulator. Hence, the value of the accumulator is set 1000 so the
function returns value 1000.
88) int i;
main(){
int t;
for ( t=4;scanf("%d",&i)-t;printf("%dn",i))
printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p
Answer:
4--0
3--1
2--2
Explanation:
Let us assume some x= scanf("%d",&i)-t the values during execution
will be,
t i x
4 0 -4
40
41. 3 1 -2
2 2 0
89) main(){
int a= 0;int b = 20;char x =1;char y =10;
if(a,b,x,y)
printf("hello");
}
Answer:
hello
Explanation:
The comma operator has associativity from left to right. Only the
rightmost value is returned and the other values are evaluated and ignored.
Thus the value of last variable y is returned to check in if. Since it is a non
zero value if becomes true so, "hello" will be printed.
90) main(){
unsigned int i;
for(i=1;i>-2;i--)
printf("c aptitude");
}
Explanation:
i is an unsigned integer. It is compared with a signed value. Since the both
types doesn't match, signed is promoted to unsigned value. The unsigned
equivalent of -2 is a huge value so condition becomes false and control
comes out of the loop.
91) In the following pgm add a stmt in the function fun such that the address of
'a' gets stored in 'j'.
main(){
int * j;
void fun(int **);
fun(&j);
}
void fun(int **k) {
int a =0;
/* add a stmt here*/
}
Answer:
*k = &a
Explanation:
The argument of the function is a pointer to a pointer.
92) What are the following notations of defining functions known as?
i. int abc(int a,float b)
{
41
42. /* some code */
}
ii. int abc(a,b)
int a; float b;
{
/* some code*/
}
Answer:
i. ANSI C notation
ii. Kernighan & Ritche notation
93) main()
{
char *p;
p="%dn";
p++;
p++;
printf(p-2,300);
}
Answer:
300
Explanation:
The pointer points to % since it is incremented twice and again
decremented by 2, it points to '%dn' and 300 is printed.
94) main(){
char a[100];
a[0]='a';a[1]]='b';a[2]='c';a[4]='d';
abc(a);
}
abc(char a[]){
a++;
printf("%c",*a);
a++;
printf("%c",*a);
}
Explanation:
The base address is modified only in function and as a result a points to 'b'
then after incrementing to 'c' so bc will be printed.
95) func(a,b)
int a,b;
{
return( a= (a==b) );
}
main()
42
43. {
int process(),func();
printf("The value of process is %d !n ",process(func,3,6));
}
process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
return((*pf) (val1,val2));
}
Answer:
The value if process is 0 !
Explanation:
The function 'process' has 3 parameters - 1, a pointer to another function 2
and 3, integers. When this function is invoked from main, the following
substitutions for formal parameters take place: func for pf, 3 for val1 and 6
for val2. This function returns the result of the operation performed by the
function 'func'. The function func has two integer parameters. The formal
parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6,
a==b returns 0. therefore the function returns 0 which in turn is returned
by the function 'process'.
96) void main()
{
static int i=5;
if(--i){
main();
printf("%d ",i);
}
}
Answer:
0 0 0 0
Explanation:
The variable "I" is declared as static, hence memory for I will be allocated
for only once, as it encounters the statement. The function main() will be called
recursively unless I becomes equal to 0, and since main() is recursively called, so
the value of static I ie., 0 will be printed every time the control is returned.
97) void main()
{
int k=ret(sizeof(float));
printf("n here value is %d",++k);
}
int ret(int ret)
{
ret += 2.5;
43
44. return(ret);
}
Answer:
Here value is 7
Explanation:
The int ret(int ret), ie., the function name and the argument name can be
the same.
Firstly, the function ret() is called in which the sizeof(float) ie., 4 is
passed, after the first expression the value in ret will be 6, as ret is integer hence
the value stored in ret will have implicit type conversion from float to int. The ret
is returned in main() it is printed after and preincrement.
98) void main()
{
char a[]="123450";
int i=strlen(a);
printf("here in 3 %dn",++i);
}
Answer:
here in 3 6
Explanation:
The char array 'a' will hold the initialized string, whose length will be
counted from 0 till the null character. Hence the 'I' will hold the value equal to 5,
after the pre-increment in the printf statement, the 6 will be printed.
99) void main()
{
unsigned giveit=-1;
int gotit;
printf("%u ",++giveit);
printf("%u n",gotit=--giveit);
}
Answer:
0 65535
Explanation:
100) void main()
{
int i;
char a[]="0";
if(printf("%sn",a))
printf("Ok here n");
else
printf("Forget itn");
}
Answer:
44
45. Ok here
Explanation:
Printf will return how many characters does it print. Hence printing
a null character returns 1 which makes the if statement true, thus
"Ok here" is printed.
101) void main()
{
void *v;
int integer=2;
int *i=&integer;
v=i;
printf("%d",(int*)*v);
}
Answer:
Compiler Error. We cannot apply indirection on type void*.
Explanation:
Void pointer is a generic pointer type. No pointer arithmetic can be
done on it. Void pointers are normally used for,
1. Passing generic pointers to functions and returning such pointers.
2. As a intermediate pointer type.
3. Used when the exact pointer type will be known at a later point of
time.
102) void main()
{
int i=i++,j=j++,k=k++;
printf(“%d%d%d”,i,j,k);
}
Answer:
Garbage values.
Explanation:
An identifier is available to use in program code from the point of its
declaration.
So expressions such as i = i++ are valid statements. The i, j and k are
automatic variables and so they contain some garbage value. Garbage in
is garbage out (GIGO).
103) void main()
{
static int i=i++, j=j++, k=k++;
printf(“i = %d j = %d k = %d”, i, j, k);
}
Answer:
i = 1 j = 1 k = 1
45
46. Explanation:
Since static variables are initialized to zero by default.
104) void main()
{
while(1){
if(printf("%d",printf("%d")))
break;
else
continue;
}
}
Answer:
Garbage values
Explanation:
The inner printf executes first to print some garbage value. The printf
returns no of characters printed and this value also cannot be predicted.
Still the outer printf prints something and so returns a non-zero value. So
it encounters the break statement and comes out of the while statement.
104) main()
{
unsigned int i=10;
while(i-->=0)
printf("%u ",i);
}
Answer:
10 9 8 7 6 5 4 3 2 1 0 65535 65534…..
Explanation:
Since i is an unsigned integer it can never become negative. So the
expression i-- >=0 will always be true, leading to an infinite loop.
105) #include<conio.h>
main()
{
int x,y=2,z,a;
if(x=y%2) z=2;
a=2;
printf("%d %d ",z,x);
}
Answer:
Garbage-value 0
Explanation:
The value of y%2 is 0. This value is assigned to x. The condition reduces
to if (x) or in other words if(0) and so z goes uninitialized.
46
47. Thumb Rule: Check all control paths to write bug free code.
106) main()
{
int a[10];
printf("%d",*a+1-*a+3);
}
Answer:
4
Explanation:
*a and -*a cancels out. The result is as simple as 1 + 3 = 4 !
107) #define prod(a,b) a*b
main()
{
int x=3,y=4;
printf("%d",prod(x+2,y-1));
}
Answer:
10
Explanation:
The macro expands and evaluates to as:
x+2*y-1 => x+(2*y)-1 => 10
108) main()
{
unsigned int i=65000;
while(i++!=0);
printf("%d",i);
}
Answer:
1
Explanation:
Note the semicolon after the while statement. When the value of i
becomes 0 it comes out of while loop. Due to post-increment on i the
value of i while printing is 1.
109) main()
{
int i=0;
while(+(+i--)!=0)
i-=i++;
printf("%d",i);
}
Answer:
-1
47
48. Explanation:
Unary + is the only dummy operator in C. So it has no effect on the
expression and now the while loop is, while(i--!=0) which is false
and so breaks out of while loop. The value –1 is printed due to the post-
decrement operator.
113) main()
{
float f=5,g=10;
enum{i=10,j=20,k=50};
printf("%dn",++k);
printf("%fn",f<<2);
printf("%lfn",f%g);
printf("%lfn",fmod(f,g));
}
Answer:
Line no 5: Error: Lvalue required
Line no 6: Cannot apply leftshift to float
Line no 7: Cannot apply mod to float
Explanation:
Enumeration constants cannot be modified, so you cannot apply ++.
Bit-wise operators and % operators cannot be applied on float values.
fmod() is to find the modulus values for floats as % operator is for ints.
110) main()
{
int i=10;
void pascal f(int,int,int);
f(i++,i++,i++);
printf(" %d",i);
}
void pascal f(integer :i,integer:j,integer :k)
{
write(i,j,k);
}
Answer:
Compiler error: unknown type integer
Compiler error: undeclared function write
Explanation:
Pascal keyword doesn’t mean that pascal code can be used. It means that
the function follows Pascal argument passing mechanism in calling the functions.
111) void pascal f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
48
49. void cdecl f(int i,int j,int k)
{
printf(“%d %d %d”,i, j, k);
}
main()
{
int i=10;
f(i++,i++,i++);
printf(" %dn",i);
i=10;
f(i++,i++,i++);
printf(" %d",i);
}
Answer:
10 11 12 13
12 11 10 13
Explanation:
Pascal argument passing mechanism forces the arguments to be called
from left to right. cdecl is the normal C argument passing mechanism where the
arguments are passed from right to left.
112). What is the output of the program given below
main()
{
signed char i=0;
for(;i>=0;i++) ;
printf("%dn",i);
}
Answer
-128
Explanation
Notice the semicolon at the end of the for loop. THe initial value of
the i is set to 0. The inner loop executes to increment the value
from 0 to 127 (the positive range of char) and then it rotates to the
negative value of -128. The condition in the for loop fails and so
comes out of the for loop. It prints the current value of i that is
-128.
113) main()
{
unsigned char i=0;
for(;i>=0;i++) ;
printf("%dn",i);
}
Answer
49
50. infinite loop
Explanation
The difference between the previous question and this one is that the char
is declared to be unsigned. So the i++ can never yield negative value and i>=0
never becomes false so that it can come out of the for loop.
114) main()
{
char i=0;
for(;i>=0;i++) ;
printf("%dn",i);
}
Answer:
Behavior is implementation dependent.
Explanation:
The detail if the char is signed/unsigned by default is
implementation dependent. If the implementation treats the char to be
signed by default the program will print –128 and terminate. On the other
hand if it considers char to be unsigned by default, it goes to infinite loop.
Rule:
You can write programs that have implementation dependent
behavior. But dont write programs that depend on such behavior.
115) Is the following statement a declaration/definition. Find what does it mean?
int (*x)[10];
Answer
Definition.
x is a pointer to array of(size 10) integers.
Apply clock-wise rule to find the meaning of this definition.
116). What is the output for the program given below
typedef enum errorType{warning, error, exception,}error;
main()
{
error g1;
g1=1;
printf("%d",g1);
}
Answer
Compiler error: Multiple declaration for error
Explanation
50
51. The name error is used in the two meanings. One means that it is a
enumerator constant with value 1. The another use is that it is a type name
(due to typedef) for enum errorType. Given a situation the compiler
cannot distinguish the meaning of error to know in what sense the error is
used:
error g1;
g1=error;
// which error it refers in each case?
When the compiler can distinguish between usages then it will not
issue error (in pure technical terms, names can only be overloaded in
different namespaces).
Note: the extra comma in the declaration,
enum errorType{warning, error, exception,}
is not an error. An extra comma is valid and is provided just for
programmer’s convenience.
117) typedef struct error{int warning, error, exception;}error;
main()
{
error g1;
g1.error =1;
printf("%d",g1.error);
}
Answer
1
Explanation
The three usages of name errors can be distinguishable by the compiler at
any instance, so valid (they are in different namespaces).
Typedef struct error{int warning, error, exception;}error;
This error can be used only by preceding the error by struct kayword as in:
struct error someError;
typedef struct error{int warning, error, exception;}error;
This can be used only after . (dot) or -> (arrow) operator preceded by the variable
name as in :
g1.error =1;
printf("%d",g1.error);
typedef struct error{int warning, error, exception;}error;
This can be used to define variables without using the preceding struct keyword
as in:
error g1;
Since the compiler can perfectly distinguish between these three usages, it is
perfectly legal and valid.
Note
51
52. This code is given here to just explain the concept behind. In real
programming don’t use such overloading of names. It reduces the readability of
the code. Possible doesn’t mean that we should use it!
118) #ifdef something
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %dn", some ,thing);
}
Answer:
Compiler error : undefined symbol some
Explanation:
This is a very simple example for conditional compilation. The
name something is not already known to the compiler making the
declaration
int some = 0;
effectively removed from the source code.
119) #if something == 0
int some=0;
#endif
main()
{
int thing = 0;
printf("%d %dn", some ,thing);
}
Answer
0 0
Explanation
This code is to show that preprocessor expressions are not the
same as the ordinary expressions. If a name is not known the
preprocessor treats it to be equal to zero.
120). What is the output for the following program
main()
{
int arr2D[3][3];
printf("%dn", ((arr2D==* arr2D)&&(* arr2D == arr2D[0])) );
52
53. }
Answer
1
Explanation
This is due to the close relation between the arrays and pointers. N
dimensional arrays are made up of (N-1) dimensional arrays.
arr2D is made up of a 3 single arrays that contains 3 integers each .
The name arr2D refers to the beginning of all the 3 arrays. *arr2D
refers to the start of the first 1D array (of 3 integers) that is the
same address as arr2D. So the expression (arr2D == *arr2D) is true
(1).
Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero
doesn’t change the value/meaning. Again arr2D[0] is the another
way of telling *(arr2D + 0). So the expression (*(arr2D + 0) ==
arr2D[0]) is true (1).
Since both parts of the expression evaluates to true the result is
true(1) and the same is printed.
121) void main()
{
if(~0 == (unsigned int)-1)
printf(“You can answer this if you know how values are represented in
memory”);
}
Answer
You can answer this if you know how values are represented in
memory
Explanation
~ (tilde operator or bit-wise negation operator) operates on 0 to
produce all ones to fill the space for an integer. –1 is represented in
unsigned value as all 1’s and so both are equal.
122) int swap(int *a,int *b)
{
*a=*a+*b;*b=*a-*b;*a=*a-*b;
}
main()
53
arr2D
arr2D[1]
arr2D[2]
arr2D[3]
54. {
int x=10,y=20;
swap(&x,&y);
printf("x= %d y = %dn",x,y);
}
Answer
x = 20 y = 10
Explanation
This is one way of swapping two values. Simple checking will help
understand this.
123) main()
{
char *p = “ayqm”;
printf(“%c”,++*(p++));
}
Answer:
b
124) main()
{
int i=5;
printf("%d",++i++);
}
Answer:
Compiler error: Lvalue required in function main
Explanation:
++i yields an rvalue. For postfix ++ to operate an lvalue is
required.
125) main()
{
char *p = “ayqm”;
char c;
c = ++*p++;
printf(“%c”,c);
}
Answer:
b
Explanation:
There is no difference between the expression ++*(p++) and +
+*p++. Parenthesis just works as a visual clue for the reader to see
which expression is first evaluated.
126)
int aaa() {printf(“Hi”);}
54
55. int bbb(){printf(“hello”);}
iny ccc(){printf(“bye”);}
main()
{
int ( * ptr[3]) ();
ptr[0] = aaa;
ptr[1] = bbb;
ptr[2] =ccc;
ptr[2]();
}
Answer:
bye
Explanation:
int (* ptr[3])() says that ptr is an array of pointers to functions that takes
no arguments and returns the type int. By the assignment ptr[0] = aaa; it
means that the first function pointer in the array is initialized with the
address of the function aaa. Similarly, the other two array elements also
get initialized with the addresses of the functions bbb and ccc. Since ptr[2]
contains the address of the function ccc, the call to the function ptr[2]() is
same as calling ccc(). So it results in printing "bye".
127)
main()
{
int i=5;
printf(“%d”,i=++i ==6);
}
Answer:
1
Explanation:
The expression can be treated as i = (++i==6), because == is of higher
precedence than = operator. In the inner expression, ++i is equal to 6
yielding true(1). Hence the result.
128) main()
{
char p[ ]="%dn";
p[1] = 'c';
printf(p,65);
}
Answer:
A
Explanation:
55
56. Due to the assignment p[1] = ‘c’ the string becomes, “%cn”. Since this
string becomes the format string for printf and ASCII value of 65 is ‘A’,
the same gets printed.
129) void ( * abc( int, void ( *def) () ) ) ();
Answer::
abc is a ptr to a function which takes 2 parameters .(a). an integer
variable.(b). a ptrto a funtion which returns void. the return type of the
function is void.
Explanation:
Apply the clock-wise rule to find the result.
130) main()
{
while (strcmp(“some”,”some0”))
printf(“Strings are not equaln”);
}
Answer:
No output
Explanation:
Ending the string constant with 0 explicitly makes no difference. So
“some” and “some0” are equivalent. So, strcmp returns 0 (false) hence
breaking out of the while loop.
131) main()
{
char str1[] = {‘s’,’o’,’m’,’e’};
char str2[] = {‘s’,’o’,’m’,’e’,’0’};
while (strcmp(str1,str2))
printf(“Strings are not equaln”);
}
Answer:
“Strings are not equal”
“Strings are not equal”
….
Explanation:
If a string constant is initialized explicitly with characters, ‘0’ is not
appended automatically to the string. Since str1 doesn’t have null
termination, it treats whatever the values that are in the following positions
as part of the string until it randomly reaches a ‘0’. So str1 and str2 are
not the same, hence the result.
132) main()
{
56
57. int i = 3;
for (;i++=0;) printf(“%d”,i);
}
Answer:
Compiler Error: Lvalue required.
Explanation:
As we know that increment operators return rvalues and hence it
cannot appear on the left hand side of an assignment operation.
133) void main()
{
int *mptr, *cptr;
mptr = (int*)malloc(sizeof(int));
printf(“%d”,*mptr);
int *cptr = (int*)calloc(sizeof(int),1);
printf(“%d”,*cptr);
}
Answer:
garbage-value 0
Explanation:
The memory space allocated by malloc is uninitialized, whereas calloc
returns the allocated memory space initialized to zeros.
134) void main()
{
static int i;
while(i<=10)
(i>2)?i++:i--;
printf(“%d”, i);
}
Answer:
32767
Explanation:
Since i is static it is initialized to 0. Inside the while loop the conditional
operator evaluates to false, executing i--. This continues till the integer
value rotates to positive value (32767). The while condition becomes false
and hence, comes out of the while loop, printing the i value.
135) main()
{
int i=10,j=20;
j = i, j?(i,j)?i:j:j;
printf("%d %d",i,j);
}
57
58. Answer:
10 10
Explanation:
The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the
question can be written as:
if(i,j)
{
if(i,j)
j = i;
else
j = j;
}
else
j = j;
136) 1. const char *a;
2. char* const a;
3. char const *a;
-Differentiate the above declarations.
Answer:
1. 'const' applies to char * rather than 'a' ( pointer to a constant char )
*a='F' : illegal
a="Hi" : legal
2. 'const' applies to 'a' rather than to the value of a (constant pointer to
char )
*a='F' : legal
a="Hi" : illegal
3. Same as 1.
137) main()
{
int i=5,j=10;
i=i&=j&&10;
printf("%d %d",i,j);
}
Answer:
1 10
Explanation:
The expression can be written as i=(i&=(j&&10)); The inner expression
(j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the
result.
58
59. 138) main()
{
int i=4,j=7;
j = j || i++ && printf("YOU CAN");
printf("%d %d", i, j);
}
Answer:
4 1
Explanation:
The boolean expression needs to be evaluated only till the truth value of
the expression is not known. j is not equal to zero itself means that the
expression’s truth value is 1. Because it is followed by || and true ||
(anything) => true where (anything) will not be evaluated. So the
remaining expression is not evaluated and so the value of i remains the
same.
Similarly when && operator is involved in an expression, when any of the
operands become false, the whole expression’s truth value becomes false
and hence the remaining expression will not be evaluated.
false && (anything) => false where (anything) will not be evaluated.
139) main()
{
register int a=2;
printf("Address of a = %d",&a);
printf("Value of a = %d",a);
}
Answer:
Compier Error: '&' on register variable
Rule to Remember:
& (address of ) operator cannot be applied on register variables.
140) main()
{
float i=1.5;
switch(i)
{
case 1: printf("1");
case 2: printf("2");
default : printf("0");
}
}
Answer:
Compiler Error: switch expression not integral
Explanation:
59
60. Switch statements can be applied only to integral types.
141) main()
{
extern i;
printf("%dn",i);
{
int i=20;
printf("%dn",i);
}
}
Answer:
Linker Error : Unresolved external symbol i
Explanation:
The identifier i is available in the inner block and so using extern has no
use in resolving it.
142) main()
{
int a=2,*f1,*f2;
f1=f2=&a;
*f2+=*f2+=a+=2.5;
printf("n%d %d %d",a,*f1,*f2);
}
Answer:
16 16 16
Explanation:
f1 and f2 both refer to the same memory location a. So changes through f1
and f2 ultimately affects only the value of a.
143) main()
{
char *p="GOOD";
char a[ ]="GOOD";
printf("n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p),
sizeof(*p), strlen(p));
printf("n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
}
Answer:
sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
sizeof(a) = 5, strlen(a) = 4
Explanation:
sizeof(p) => sizeof(char*) => 2
sizeof(*p) => sizeof(char) => 1
Similarly,
sizeof(a) => size of the character array => 5
60
61. When sizeof operator is applied to an array it returns the sizeof the array
and it is not the same as the sizeof the pointer variable. Here the sizeof(a)
where a is the character array and the size of the array is 5 because the
space necessary for the terminating NULL character should also be taken
into account.
144) #define DIM( array, type) sizeof(array)/sizeof(type)
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr, int));
}
Answer:
10
Explanation:
The size of integer array of 10 elements is 10 * sizeof(int). The macro
expands to sizeof(arr)/sizeof(int) => 10 * sizeof(int) / sizeof(int) => 10.
145) int DIM(int array[])
{
return sizeof(array)/sizeof(int );
}
main()
{
int arr[10];
printf(“The dimension of the array is %d”, DIM(arr));
}
Answer:
1
Explanation:
Arrays cannot be passed to functions as arguments and only the pointers
can be passed. So the argument is equivalent to int * array (this is one of
the very few places where [] and * usage are equivalent). The return
statement becomes, sizeof(int *)/ sizeof(int) that happens to be equal in
this case.
146) main()
{
static int a[3][3]={1,2,3,4,5,6,7,8,9};
int i,j;
static *p[]={a,a+1,a+2};
for(i=0;i<3;i++)
{
for(j=0;j<3;j++)
printf("%dt%dt%dt%dn",*(*(p+i)+j),
61
62. *(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
}
}
Answer:
1 1 1 1
2 4 2 4
3 7 3 7
4 2 4 2
5 5 5 5
6 8 6 8
7 3 7 3
8 6 8 6
9 9 9 9
Explanation:
*(*(p+i)+j) is equivalent to p[i][j].
147) main()
{
void swap();
int x=10,y=8;
swap(&x,&y);
printf("x=%d y=%d",x,y);
}
void swap(int *a, int *b)
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
Answer:
x=10 y=8
Explanation:
Using ^ like this is a way to swap two variables without using a temporary
variable and that too in a single statement.
Inside main(), void swap(); means that swap is a function that may take
any number of arguments (not no arguments) and returns nothing. So this
doesn’t issue a compiler error by the call swap(&x,&y); that has two
arguments.
This convention is historically due to pre-ANSI style (referred to as
Kernighan and Ritchie style) style of function declaration. In that style, the
swap function will be defined as follows,
void swap()
int *a, int *b
{
*a ^= *b, *b ^= *a, *a ^= *b;
}
62
63. where the arguments follow the (). So naturally the declaration for swap
will look like, void swap() which means the swap can take any number of
arguments.
148) main()
{
int i = 257;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
1 1
Explanation:
The integer value 257 is stored in the memory as, 00000001 00000001, so
the individual bytes are taken by casting it to char * and get printed.
149) main()
{
int i = 258;
int *iPtr = &i;
printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}
Answer:
2 1
Explanation:
The integer value 257 can be represented in binary as, 00000001
00000001. Remember that the INTEL machines are ‘small-endian’
machines. Small-endian means that the lower order bytes are stored in the
higher memory addresses and the higher order bytes are stored in lower
addresses. The integer value 258 is stored in memory as: 00000001
00000010.
150) main()
{
int i=300;
char *ptr = &i;
*++ptr=2;
printf("%d",i);
}
Answer:
556
Explanation:
The integer value 300 in binary notation is: 00000001 00101100. It is
stored in memory (small-endian) as: 00101100 00000001. Result of the
expression *++ptr = 2 makes the memory representation as: 00101100
63
64. 00000010. So the integer corresponding to it is 00000010 00101100 =>
556.
151) #include <stdio.h>
main()
{
char * str = "hello";
char * ptr = str;
char least = 127;
while (*ptr++)
least = (*ptr<least ) ?*ptr :least;
printf("%d",least);
}
Answer:
0
Explanation:
After ‘ptr’ reaches the end of the string the value pointed by ‘str’ is ‘0’.
So the value of ‘str’ is less than that of ‘least’. So the value of ‘least’
finally is 0.
152) Declare an array of N pointers to functions returning pointers to functions
returning pointers to characters?
Answer:
(char*(*)( )) (*ptr[N])( );
153) main()
{
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno, &student.dob.day,
&student.dob.month, &student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Inside the struct definition of ‘student’ the member of type struct date is
given. The compiler doesn’t have the definition of date structure (forward
reference is not allowed in C in this case) so it issues an error.
64
65. 154) main()
{
struct date;
struct student
{
char name[30];
struct date dob;
}stud;
struct date
{
int day,month,year;
};
scanf("%s%d%d%d", stud.rollno, &student.dob.day, &student.dob.month,
&student.dob.year);
}
Answer:
Compiler Error: Undefined structure date
Explanation:
Only declaration of struct date is available inside the structure definition
of ‘student’ but to have a variable of type struct date the definition of the
structure is required.
155) There were 10 records stored in “somefile.dat” but the following program printed
11 names. What went wrong?
void main()
{
struct student
{
char name[30], rollno[6];
}stud;
FILE *fp = fopen(“somefile.dat”,”r”);
while(!feof(fp))
{
fread(&stud, sizeof(stud), 1 , fp);
puts(stud.name);
}
}
Explanation:
fread reads 10 records and prints the names successfully. It will
return EOF only when fread tries to read another record and fails
reading EOF (and returning EOF). So it prints the last record
again. After this only the condition feof(fp) becomes false, hence
comes out of the while loop.
156) Is there any difference between the two declarations,
1. int foo(int *arr[]) and
65
66. 2. int foo(int *arr[2])
Answer:
No
Explanation:
Functions can only pass pointers and not arrays. The numbers that are
allowed inside the [] is just for more readability. So there is no difference
between the two declarations.
157) What is the subtle error in the following code segment?
void fun(int n, int arr[])
{
int *p=0;
int i=0;
while(i++<n)
p = &arr[i];
*p = 0;
}
Answer & Explanation:
If the body of the loop never executes p is assigned no address. So
p remains NULL where *p =0 may result in problem (may rise to
runtime error “NULL pointer assignment” and terminate the
program).
158) What is wrong with the following code?
int *foo()
{
int *s = malloc(sizeof(int)100);
assert(s != NULL);
return s;
}
Answer & Explanation:
assert macro should be used for debugging and finding out bugs. The
check s != NULL is for error/exception handling and for that assert
shouldn’t be used. A plain if and the corresponding remedy statement has
to be given.
159) What is the hidden bug with the following statement?
assert(val++ != 0);
Answer & Explanation:
Assert macro is used for debugging and removed in release version. In
assert, the experssion involves side-effects. So the behavior of the code
becomes different in case of debug version and the release version thus
leading to a subtle bug.
Rule to Remember:
Don’t use expressions that have side-effects in assert statements.
66
67. 160) void main()
{
int *i = 0x400; // i points to the address 400
*i = 0; // set the value of memory location pointed by i;
}
Answer:
Undefined behavior
Explanation:
The second statement results in undefined behavior because it points to
some location whose value may not be available for modification. This
type of pointer in which the non-availability of the implementation of the
referenced location is known as 'incomplete type'.
161) #define assert(cond) if(!(cond))
(fprintf(stderr, "assertion failed: %s, file %s, line %d n",#cond,
__FILE__,__LINE__), abort())
void main()
{
int i = 10;
if(i==0)
assert(i < 100);
else
printf("This statement becomes else for if in assert macro");
}
Answer:
No output
Explanation:
The else part in which the printf is there becomes the else for if in the assert
macro. Hence nothing is printed.
The solution is to use conditional operator instead of if statement,
#define assert(cond) ((cond)?(0): (fprintf (stderr, "assertion failed: %s, file %s,
line %d n",#cond, __FILE__,__LINE__), abort()))
Note:
However this problem of “matching with nearest else” cannot be solved
by the usual method of placing the if statement inside a block like this,
#define assert(cond) {
if(!(cond))
(fprintf(stderr, "assertion failed: %s, file %s, line %d n",#cond,
__FILE__,__LINE__), abort())
}
162) Is the following code legal?
struct a
67
68. {
int x;
struct a b;
}
Answer:
No
Explanation:
Is it not legal for a structure to contain a member that is of the same
type as in this case. Because this will cause the structure declaration to be
recursive without end.
163) Is the following code legal?
struct a
{
int x;
struct a *b;
}
Answer:
Yes.
Explanation:
*b is a pointer to type struct a and so is legal. The compiler knows, the
size of the pointer to a structure even before the size of the structure
is determined(as you know the pointer to any type is of same size). This
type of structures is known as ‘self-referencing’ structure.
164) Is the following code legal?
typedef struct a
{
int x;
aType *b;
}aType
Answer:
No
Explanation:
The typename aType is not known at the point of declaring the structure
(forward references are not made for typedefs).
165) Is the following code legal?
typedef struct a aType;
struct a
{
int x;
aType *b;
};
Answer:
Yes
68
69. Explanation:
The typename aType is known at the point of declaring the structure,
because it is already typedefined.
166) Is the following code legal?
void main()
{
typedef struct a aType;
aType someVariable;
struct a
{
int x;
aType *b;
};
}
Answer:
No
Explanation:
When the declaration,
typedef struct a aType;
is encountered body of struct a is not known. This is known as ‘incomplete
types’.
167) void main()
{
printf(“sizeof (void *) = %d n“, sizeof( void *));
printf(“sizeof (int *) = %d n”, sizeof(int *));
printf(“sizeof (double *) = %d n”, sizeof(double *));
printf(“sizeof(struct unknown *) = %d n”, sizeof(struct unknown *));
}
Answer :
sizeof (void *) = 2
sizeof (int *) = 2
sizeof (double *) = 2
sizeof(struct unknown *) = 2
Explanation:
The pointer to any type is of same size.
168) char inputString[100] = {0};
To get string input from the keyboard which one of the following is better?
1) gets(inputString)
2) fgets(inputString, sizeof(inputString), fp)
Answer & Explanation:
The second one is better because gets(inputString) doesn't know the size
of the string passed and so, if a very big input (here, more than 100 chars)
69
70. the charactes will be written past the input string. When fgets is used with
stdin performs the same operation as gets but is safe.
169) Which version do you prefer of the following two,
1) printf(“%s”,str); // or the more curt one
2) printf(str);
Answer & Explanation:
Prefer the first one. If the str contains any format characters like %d then
it will result in a subtle bug.
170) void main()
{
int i=10, j=2;
int *ip= &i, *jp = &j;
int k = *ip/*jp;
printf(“%d”,k);
}
Answer:
Compiler Error: “Unexpected end of file in comment started in line 5”.
Explanation:
The programmer intended to divide two integers, but by the
“maximum munch” rule, the compiler treats the operator
sequence / and * as /* which happens to be the starting of
comment. To force what is intended by the programmer,
int k = *ip/ *jp;
// give space explicity separating / and *
//or
int k = *ip/(*jp);
// put braces to force the intention
will solve the problem.
171) void main()
{
char ch;
for(ch=0;ch<=127;ch++)
printf(“%c %d n“, ch, ch);
}
Answer:
Implementaion dependent
Explanation:
The char type may be signed or unsigned by default. If it is signed then
ch++ is executed after ch reaches 127 and rotates back to -128. Thus ch is
always smaller than 127.
172) Is this code legal?
int *ptr;
70
71. ptr = (int *) 0x400;
Answer:
Yes
Explanation:
The pointer ptr will point at the integer in the memory location 0x400.
173) main()
{
char a[4]="HELLO";
printf("%s",a);
}
Answer:
Compiler error: Too many initializers
Explanation:
The array a is of size 4 but the string constant requires 6 bytes to get
stored.
174) main()
{
char a[4]="HELL";
printf("%s",a);
}
Answer:
HELL%@!~@!@???@~~!
Explanation:
The character array has the memory just enough to hold the string
“HELL” and doesnt have enough space to store the terminating null
character. So it prints the HELL correctly and continues to print garbage
values till it accidentally comes across a NULL character.
175) main()
{
int a=10,*j;
void *k;
j=k=&a;
j++;
k++;
printf("n %u %u ",j,k);
}
Answer:
Compiler error: Cannot increment a void pointer
Explanation:
Void pointers are generic pointers and they can be used only when the
type is not known and as an intermediate address storage type. No pointer
arithmetic can be done on it and you cannot apply indirection operator (*)
on void pointers.
71
72. 176) main()
{
extern int i;
{ int i=20;
{
const volatile unsigned i=30; printf("%d",i);
}
printf("%d",i);
}
printf("%d",i);
}
int i;
177) Printf can be implemented by using __________ list.
Answer:
Variable length argument lists
178) char *someFun()
{
char *temp = “string constant";
return temp;
}
int main()
{
puts(someFun());
}
Answer:
string constant
Explanation:
The program suffers no problem and gives the output correctly because the
character constants are stored in code/data area and not allocated in stack, so this doesn’t
lead to dangling pointers.
179) char *someFun1()
{
char temp[ ] = “string";
return temp;
}
char *someFun2()
{
char temp[ ] = {‘s’, ‘t’,’r’,’i’,’n’,’g’};
return temp;
}
int main()
{
puts(someFun1());
72
73. puts(someFun2());
}
Answer:
Garbage values.
Explanation:
Both the functions suffer from the problem of dangling pointers. In someFun1()
temp is a character array and so the space for it is allocated in heap and is initialized with
character string “string”. This is created dynamically as the function is called, so is also
deleted dynamically on exiting the function so the string data is not available in the
calling function main() leading to print some garbage values. The function someFun2()
also suffers from the same problem but the problem can be easily identified in this case.
73
74. C++ Aptitude and OOPS
Note : All the programs are tested under Turbo C++ 3.0, 4.5 and Microsoft VC++ 6.0
compilers.
It is assumed that,
Programs run under Windows environment,
The underlying machine is an x86 based system,
Program is compiled using Turbo C/C++ compiler.
The program output may depend on the information based on this assumptions
(for example sizeof(int) == 2 may be assumed).
1) class Sample
{
public:
int *ptr;
Sample(int i)
{
ptr = new int(i);
}
~Sample()
{
delete ptr;
}
void PrintVal()
{
cout << "The value is " << *ptr;
}
};
void SomeFunc(Sample x)
{
cout << "Say i am in someFunc " << endl;
}
int main()
{
Sample s1= 10;
SomeFunc(s1);
s1.PrintVal();
}
Answer:
74
C++ Aptitude
and OOPS
75. Say i am in someFunc
Null pointer assignment(Run-time error)
Explanation:
As the object is passed by value to SomeFunc the destructor of the object is
called when the control returns from the function. So when PrintVal is called it meets up
with ptr that has been freed.The solution is to pass the Sample object by reference to
SomeFunc:
void SomeFunc(Sample &x)
{
cout << "Say i am in someFunc " << endl;
}
because when we pass objects by refernece that object is not destroyed. while returning
from the function.
2) Which is the parameter that is added to every non-static member function when it is
called?
Answer:
‘this’ pointer
3) class base
{
public:
int bval;
base(){ bval=0;}
};
class deri:public base
{
public:
int dval;
deri(){ dval=1;}
};
void SomeFunc(base *arr,int size)
{
for(int i=0; i<size; i++,arr++)
cout<<arr->bval;
cout<<endl;
}
int main()
{
base BaseArr[5];
SomeFunc(BaseArr,5);
deri DeriArr[5];
SomeFunc(DeriArr,5);
75
76. }
Answer:
00000
01010
Explanation:
The function SomeFunc expects two arguments.The first one is a pointer to an
array of base class objects and the second one is the sizeof the array.The first call of
someFunc calls it with an array of bae objects, so it works correctly and prints the bval of
all the objects. When Somefunc is called the second time the argument passed is the
pointeer to an array of derived class objects and not the array of base class objects. But
that is what the function expects to be sent. So the derived class pointer is promoted to
base class pointer and the address is sent to the function. SomeFunc() knows nothing
about this and just treats the pointer as an array of base class objects. So when arr++ is
met, the size of base class object is taken into consideration and is incremented by
sizeof(int) bytes for bval (the deri class objects have bval and dval as members and so is
of size >= sizeof(int)+sizeof(int) ).
4) class base
{
public:
void baseFun(){ cout<<"from base"<<endl;}
};
class deri:public base
{
public:
void baseFun(){ cout<< "from derived"<<endl;}
};
void SomeFunc(base *baseObj)
{
baseObj->baseFun();
}
int main()
{
base baseObject;
SomeFunc(&baseObject);
deri deriObject;
SomeFunc(&deriObject);
}
Answer:
from base
from base
Explanation:
As we have seen in the previous case, SomeFunc expects a pointer to a base class.
Since a pointer to a derived class object is passed, it treats the argument only as a base
class pointer and the corresponding base function is called.
76
77. 5) class base
{
public:
virtual void baseFun(){ cout<<"from base"<<endl;}
};
class deri:public base
{
public:
void baseFun(){ cout<< "from derived"<<endl;}
};
void SomeFunc(base *baseObj)
{
baseObj->baseFun();
}
int main()
{
base baseObject;
SomeFunc(&baseObject);
deri deriObject;
SomeFunc(&deriObject);
}
Answer:
from base
from derived
Explanation:
Remember that baseFunc is a virtual function. That means that it supports run-
time polymorphism. So the function corresponding to the derived class object is called.
void main()
{
int a, *pa, &ra;
pa = &a;
ra = a;
cout <<"a="<<a <<"*pa="<<*pa <<"ra"<<ra ;
}
/*
Answer :
Compiler Error: 'ra',reference must be initialized
Explanation :
Pointers are different from references. One of the main
differences is that the pointers can be both initialized and assigned,
whereas references can only be initialized. So this code issues an error.
*/
77
78. const int size = 5;
void print(int *ptr)
{
cout<<ptr[0];
}
void print(int ptr[size])
{
cout<<ptr[0];
}
void main()
{
int a[size] = {1,2,3,4,5};
int *b = new int(size);
print(a);
print(b);
}
/*
Answer:
Compiler Error : function 'void print(int *)' already has a body
Explanation:
Arrays cannot be passed to functions, only pointers (for arrays, base addresses)
can be passed. So the arguments int *ptr and int prt[size] have no difference
as function arguments. In other words, both the functoins have the same signature and
so cannot be overloaded.
*/
class some{
public:
~some()
{
cout<<"some's destructor"<<endl;
}
};
void main()
{
some s;
s.~some();
}
/*
Answer:
some's destructor
some's destructor
78
79. Explanation:
Destructors can be called explicitly. Here 's.~some()' explicitly calls the
destructor of 's'. When main() returns, destructor of s is called again,
hence the result.
*/
#include <iostream.h>
class fig2d
{
int dim1;
int dim2;
public:
fig2d() { dim1=5; dim2=6;}
virtual void operator<<(ostream & rhs);
};
void fig2d::operator<<(ostream &rhs)
{
rhs <<this->dim1<<" "<<this->dim2<<" ";
}
/*class fig3d : public fig2d
{
int dim3;
public:
fig3d() { dim3=7;}
virtual void operator<<(ostream &rhs);
};
void fig3d::operator<<(ostream &rhs)
{
fig2d::operator <<(rhs);
rhs<<this->dim3;
}
*/
void main()
{
fig2d obj1;
// fig3d obj2;
obj1 << cout;
// obj2 << cout;
}
79
80. /*
Answer :
5 6
Explanation:
In this program, the << operator is overloaded with ostream as argument.
This enables the 'cout' to be present at the right-hand-side. Normally, 'cout'
is implemented as global function, but it doesn't mean that 'cout' is not possible
to be overloaded as member function.
Overloading << as virtual member function becomes handy when the class in which
it is overloaded is inherited, and this becomes available to be overrided. This is as
opposed
to global friend functions, where friend's are not inherited.
*/
class opOverload{
public:
bool operator==(opOverload temp);
};
bool opOverload::operator==(opOverload temp){
if(*this == temp ){
cout<<"The both are same objectsn";
return true;
}
else{
cout<<"The both are differentn";
return false;
}
}
void main(){
opOverload a1, a2;
a1= =a2;
}
Answer :
Runtime Error: Stack Overflow
Explanation :
Just like normal functions, operator functions can be called recursively. This
program just illustrates that point, by calling the operator == function recursively, leading
to an infinite loop.
class complex{
double re;
double im;
80
81. public:
complex() : re(1),im(0.5) {}
bool operator==(complex &rhs);
operator int(){}
};
bool complex::operator == (complex &rhs){
if((this->re == rhs.re) && (this->im == rhs.im))
return true;
else
return false;
}
int main(){
complex c1;
cout<< c1;
}
Answer : Garbage value
Explanation:
The programmer wishes to print the complex object using output
re-direction operator,which he has not defined for his lass.But the compiler instead of
giving an error sees the conversion function
and converts the user defined object to standard object and prints
some garbage value.
class complex{
double re;
double im;
public:
complex() : re(0),im(0) {}
complex(double n) { re=n,im=n;};
complex(int m,int n) { re=m,im=n;}
void print() { cout<<re; cout<<im;}
};
void main(){
complex c3;
double i=5;
c3 = i;
c3.print();
}
Answer:
81
82. 5,5
Explanation:
Though no operator= function taking complex, double is defined, the double on
the rhs is converted into a temporary object using the single argument constructor taking
double and assigned to the lvalue.
void main()
{
int a, *pa, &ra;
pa = &a;
ra = a;
cout <<"a="<<a <<"*pa="<<*pa <<"ra"<<ra ;
}
Answer :
Compiler Error: 'ra',reference must be initialized
Explanation :
Pointers are different from references. One of the main
differences is that the pointers can be both initialized and assigned,
whereas references can only be initialized. So this code issues an error.
Try it Yourself
1) Determine the output of the 'C++' Codelet.
class base
{
public :
out()
{
cout<<"base ";
}
};
class deri{
public : out()
{
cout<<"deri ";
}
};
void main()
{ deri dp[3];
base *bp = (base*)dp;
for (int i=0; i<3;i++)
(bp++)->out();
}
82
83. 2) Justify the use of virtual constructors and destructors in C++.
3) Each C++ object possesses the 4 member fns,(which can be declared by the
programmer explicitly or by the implementation if they are not available). What are
those 4 functions?
4) What is wrong with this class declaration?
class something
{
char *str;
public:
something(){
st = new char[10]; }
~something()
{
delete str;
}
};
5) Inheritance is also known as -------- relationship. Containership as ________
relationship.
6) When is it necessary to use member-wise initialization list (also known as header
initialization list) in C++?
7) Which is the only operator in C++ which can be overloaded but NOT inherited.
8) Is there anything wrong with this C++ class declaration?
class temp
{
int value1;
mutable int value2;
public :
void fun(int val)
const{
((temp*) this)->value1 = 10;
value2 = 10;
}
};
83
84. 1. What is a modifier?
Answer:
A modifier, also called a modifying function is a member function that changes the
value of at least one data member. In other words, an operation that modifies the state of
an object. Modifiers are also known as ‘mutators’.
2. What is an accessor?
Answer:
An accessor is a class operation that does not modify the state of an object. The
accessor functions need to be declared as const operations
3. Differentiate between a template class and class template.
Answer:
Template class:
A generic definition or a parameterized class not instantiated until the client
provides the needed information. It’s jargon for plain templates.
Class template:
A class template specifies how individual classes can be constructed much like
the way a class specifies how individual objects can be constructed. It’s jargon for plain
classes.
4. When does a name clash occur?
Answer:
A name clash occurs when a name is defined in more than one place. For
example., two different class libraries could give two different classes the same name. If
you try to use many class libraries at the same time, there is a fair chance that you will be
unable to compile or link the program because of name clashes.
5. Define namespace.
Answer:
It is a feature in c++ to minimize name collisions in the global name space. This
namespace keyword assigns a distinct name to a library that allows other libraries to use
the same identifier names without creating any name collisions. Furthermore, the
compiler uses the namespace signature for differentiating the definitions.
6. What is the use of ‘using’ declaration.
Answer:
A using declaration makes it possible to use a name from a namespace without the
scope operator.
7. What is an Iterator class?
Answer:
A class that is used to traverse through the objects maintained by a container
class. There are five categories of iterators:
input iterators,
output iterators,
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85. forward iterators,
bidirectional iterators,
random access.
An iterator is an entity that gives access to the contents of a container object
without violating encapsulation constraints. Access to the contents is granted on a one-at-
a-time basis in order. The order can be storage order (as in lists and queues) or some
arbitrary order (as in array indices) or according to some ordering relation (as in an
ordered binary tree). The iterator is a construct, which provides an interface that, when
called, yields either the next element in the container, or some value denoting the fact that
there are no more elements to examine. Iterators hide the details of access to and update
of the elements of a container class.
The simplest and safest iterators are those that permit read-only access to the
contents of a container class. The following code fragment shows how an iterator might
appear in code:
cont_iter:=new cont_iterator();
x:=cont_iter.next();
while x/=none do
...
s(x);
...
x:=cont_iter.next();
end;
In this example, cont_iter is the name of the iterator. It is created on the first line by
instantiation of cont_iterator class, an iterator class defined to iterate over some container
class, cont. Succesive elements from the container are carried to x. The loop terminates
when x is bound to some empty value. (Here, none)In the middle of the loop, there is s(x)
an operation on x, the current element from the container. The next element of the
container is obtained at the bottom of the loop.
9. List out some of the OODBMS available.
Answer:
GEMSTONE/OPAL of Gemstone systems.
ONTOS of Ontos.
Objectivity of Objectivity inc.
Versant of Versant object technology.
Object store of Object Design.
ARDENT of ARDENT software.
POET of POET software.
10. List out some of the object-oriented methodologies.
Answer:
Object Oriented Development (OOD) (Booch 1991,1994).
Object Oriented Analysis and Design (OOA/D) (Coad and Yourdon 1991).
Object Modelling Techniques (OMT) (Rumbaugh 1991).
Object Oriented Software Engineering (Objectory) (Jacobson 1992).
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86. Object Oriented Analysis (OOA) (Shlaer and Mellor 1992).
The Fusion Method (Coleman 1991).
11. What is an incomplete type?
Answer:
Incomplete types refers to pointers in which there is non availability of the
implementation of the referenced location or it points to some location whose value is not
available for modification.
Example:
int *i=0x400 // i points to address 400
*i=0; //set the value of memory location pointed by i.
Incomplete types are otherwise called uninitialized pointers.
12. What is a dangling pointer?
Answer:
A dangling pointer arises when you use the address of an object after its lifetime
is over.
This may occur in situations like returning addresses of the automatic variables from a
function or using the address of the memory block after it is freed.
13. Differentiate between the message and method.
Answer:
Message Method
Objects communicate by sending messages Provides response to a message.
to each other.
A message is sent to invoke a method. It is an implementation of an operation.
14. What is an adaptor class or Wrapper class?
Answer:
A class that has no functionality of its own. Its member functions hide the use of a
third party software component or an object with the non-compatible interface or a non-
object- oriented implementation.
15. What is a Null object?
Answer:
It is an object of some class whose purpose is to indicate that a real object of that
class does not exist. One common use for a null object is a return value from a member
function that is supposed to return an object with some specified properties but cannot
find such an object.
16. What is class invariant?
Answer:
A class invariant is a condition that defines all valid states for an object. It is a
logical condition to ensure the correct working of a class. Class invariants must hold
when an object is created, and they must be preserved under all operations of the class. In
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87. particular all class invariants are both preconditions and post-conditions for all operations
or member functions of the class.
17. What do you mean by Stack unwinding?
Answer:
It is a process during exception handling when the destructor is called for all local
objects between the place where the exception was thrown and where it is caught.
18. Define precondition and post-condition to a member function.
Answer:
Precondition:
A precondition is a condition that must be true on entry to a member function. A
class is used correctly if preconditions are never false. An operation is not responsible for
doing anything sensible if its precondition fails to hold.
For example, the interface invariants of stack class say nothing about pushing yet
another element on a stack that is already full. We say that isful() is a precondition of the
push operation.
Post-condition:
A post-condition is a condition that must be true on exit from a member function
if the precondition was valid on entry to that function. A class is implemented correctly if
post-conditions are never false.
For example, after pushing an element on the stack, we know that isempty() must
necessarily hold. This is a post-condition of the push operation.
19. What are the conditions that have to be met for a condition to be an invariant of the
class?
Answer:
The condition should hold at the end of every constructor.
The condition should hold at the end of every mutator(non-const) operation.
20. What are proxy objects?
Answer:
Objects that stand for other objects are called proxy objects or surrogates.
Example:
template<class T>
class Array2D
{
public:
class Array1D
{
public:
T& operator[] (int index);
const T& operator[] (int index) const;
...
};
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