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C PROGRAMING 202 - 2024



Section-A:

1. What are header files and their uses?

Header files in C contain declarations of functions, macros, constants, and data types used in programs. They allow code reuse and modular programming by providing predefined functions and simplifying complex tasks.

Common Header Files:

  • #include <stdio.h> – Standard Input/Output functions (e.g., printf(), scanf()).
  • #include <stdlib.h> – Memory allocation and process control (malloc(), exit()).
  • #include <string.h> – String handling functions (strlen(), strcpy()).

Uses:

  • Code Reusability: Avoids rewriting common code.
  • Modularity: Organizes code logically.
  • Simplifies Complex Operations: Provides ready-to-use functions.

2. What is the difference between macro and functions?

Feature

Macro

Function

Definition

Defined using #define preprocessor directive.

Defined using the return_type and function_name.

Execution

Replaced by code during compilation.

Called during program execution.

Speed

Faster (no function call overhead).

Slower due to function call overhead.

Debugging

Harder to debug (no type checking).

Easier to debug (type checking enforced).

Example

#define SQUARE(x) ((x)*(x))

int square(int x) { return x*x; }

3. What is a structure? How is it different from Union?

A structure (struct) in C is a user-defined data type that allows grouping variables of different types under one name.

Structure Example:

struct Student {

    int id;

    char name[20];

    float marks;

};

A union also groups variables of different types but shares the same memory location for all members.

Union Example:

union Data {

    int i;

    float f;

    char str[20];

};

Difference Between Structure and Union:

Feature

Structure

Union

Memory

Allocates memory for all members.

Shares memory among members.

Usage

All members can be used simultaneously.

Only one member holds a value at a time.

Size

Sum of all members' sizes.

Size of the largest member.

4. What is static memory allocation and dynamic memory allocation?

Static Memory Allocation:

  • Memory is allocated during compile-time.
  • The size is fixed and cannot be changed during runtime.
  • Faster and safer.
  • Example:
  • int arr[10];  // Static array of size 10

Dynamic Memory Allocation:

  • Memory is allocated during runtime.
  • The size can change as needed.
  • Managed using functions like malloc(), calloc(), realloc(), and free().
  • Example:
  • int *ptr;
  • ptr = (int *)malloc(10 * sizeof(int));  // Dynamic array of size 10

Difference:

Feature

Static Memory Allocation

Dynamic Memory Allocation

When Allocated

Compile-time

Runtime

Memory Size

Fixed

Flexible

Functions Used

None

malloc(), calloc()

Efficiency

Faster

Slightly slower

5. What is the difference between getc(), getchar(), getch(), and getche()?

Function

Purpose

Echo Input?

Buffered?

Header File

getc()

Reads a character from a file or stdin.

Yes

Yes

stdio.h

getchar()

Reads a character from standard input.

Yes

Yes

stdio.h

getch()

Reads a character without echoing input.

No

No

conio.h

getche()

Reads a character and echoes input.

Yes

No

conio.h

Examples:

char ch;

// Using getc()

ch = getc(stdin); 

 

// Using getchar()

ch = getchar();  

 

// Using getch()

ch = getch();   

 

// Using getche()

ch = getche();  

Section-B Detailed Solutions

Question 6: Write a C program that reads 20 integer numbers from keywords and sorts them in ascending order.

Explanation:
This program takes 20 integer inputs from the user, stores them in an array, and sorts them using the Bubble Sort algorithm. Bubble Sort repeatedly steps through the list, compares adjacent elements, and swaps them if they are in the wrong order.

C Program:

#include <stdio.h>

 

void bubbleSort(int arr[], int n) {

    int i, j, temp;

    for (i = 0; i < n - 1; i++) {

        for (j = 0; j < n - i - 1; j++) {

            if (arr[j] > arr[j + 1]) {

                temp = arr[j];

                arr[j] = arr[j + 1];

                arr[j + 1] = temp;

            }

        }

    }

}

 

int main() {

    int arr[20], i;

 

    printf("Enter 20 integers:\n");

    for (i = 0; i < 20; i++) {

        scanf("%d", &arr[i]);

    }

 

    bubbleSort(arr, 20);

 

    printf("Sorted numbers in ascending order:\n");

    for (i = 0; i < 20; i++) {

        printf("%d ", arr[i]);

    }

    printf("\n");

   

    return 0;

}

Explanation of the Program:

  1. Input: The program uses a loop to take 20 integer inputs.
  2. Sorting: The bubbleSort function sorts the array in ascending order.
  3. Output: The sorted array is printed.

Time Complexity: O(n^2)
Space Complexity: O(1)

Question 7: Write a C program to check input string is palindrome or not without using any string library function.

Explanation:
A palindrome is a string that reads the same backward as forward. This program reads a string from the user and checks if it is a palindrome without using string library functions.

C Program:

#include <stdio.h>

 

int stringLength(char str[]) {

    int length = 0;

    while (str[length] != '\0') {

        length++;

    }

    return length;

}

 

int isPalindrome(char str[]) {

    int i, len = stringLength(str);

    for (i = 0; i < len / 2; i++) {

        if (str[i] != str[len - i - 1]) {

            return 0; // Not a palindrome

        }

    }

    return 1; // Palindrome

}

 

int main() {

    char str[100];

 

    printf("Enter a string: ");

    scanf("%s", str);

 

    if (isPalindrome(str)) {

        printf("The string is a palindrome.\n");

    } else {

        printf("The string is not a palindrome.\n");

    }

   

    return 0;

}

Explanation of the Program:

  1. Input: Reads a string from the user.
  2. Length Calculation: The stringLength function calculates the length of the string.
  3. Palindrome Check: Compares characters from both ends moving towards the center.
  4. Output: Displays whether the string is a palindrome.

Example Output:

Enter a string: madam

The string is a palindrome.

 

Enter a string: hello

The string is not a palindrome.

Time Complexity: O(n)
Space Complexity: O(1)

Section-C Detailed Solutions

Question 8: Write a C program to add two complex numbers by passing structure to a function.

Explanation:
A complex number has a real and an imaginary part. This program defines a structure for complex numbers and uses a function to add two complex numbers.

C Program:

#include <stdio.h>

 

typedef struct {

    float real;

    float imag;

} Complex;

 

Complex addComplex(Complex a, Complex b) {

    Complex result;

    result.real = a.real + b.real;

    result.imag = a.imag + b.imag;

    return result;

}

 

int main() {

    Complex num1, num2, sum;

 

    printf("Enter real and imaginary parts of first complex number: ");

    scanf("%f %f", &num1.real, &num1.imag);

 

    printf("Enter real and imaginary parts of second complex number: ");

    scanf("%f %f", &num2.real, &num2.imag);

 

    sum = addComplex(num1, num2);

 

    printf("Sum = %.2f + %.2fi\n", sum.real, sum.imag);

 

    return 0;

}

Explanation of the Program:

  1. Defines a Complex structure with real and imaginary parts.
  2. addComplex function adds two complex numbers.
  3. Reads two complex numbers from the user and prints their sum.

Question 9: What do you mean by pointer? Write a C program to access elements in a 2-dimensional array using a pointer.

Explanation:
A pointer is a variable that stores the address of another variable. This program demonstrates accessing a 2D array using pointers.

C Program:

#include <stdio.h>

 

int main() {

    int arr[3][3], *ptr, i, j;

 

    printf("Enter elements of 3x3 matrix:\n");

    for (i = 0; i < 3; i++) {

        for (j = 0; j < 3; j++) {

            scanf("%d", &arr[i][j]);

        }

    }

 

    ptr = &arr[0][0];

 

    printf("Matrix elements:\n");

    for (i = 0; i < 3; i++) {

        for (j = 0; j < 3; j++) {

            printf("%d ", *(ptr + i * 3 + j));

        }

        printf("\n");

    }

 

    return 0;

}

Explanation of the Program:

  1. Reads a 3x3 matrix from the user.
  2. Uses a pointer to access and print matrix elements.

Question 10: How string is different from an array? Write a C program to convert a lowercase string to uppercase without using string manipulation functions.

Explanation:
Strings are arrays of characters ending with a null character \0. This program converts lowercase letters to uppercase manually.

C Program:

#include <stdio.h>

 

void toUpperCase(char str[]) {

    int i = 0;

    while (str[i] != '\0') {

        if (str[i] >= 'a' && str[i] <= 'z') {

            str[i] = str[i] - 32;

        }

        i++;

    }

}

 

int main() {

    char str[100];

 

    printf("Enter a lowercase string: ");

    scanf("%s", str);

 

    toUpperCase(str);

    printf("Uppercase String: %s\n", str);

 

    return 0;

}

Explanation of the Program:

  1. Reads a lowercase string.
  2. Converts each character to uppercase by subtracting 32 from its ASCII value.

Question 11: Write a program in C to count the number of words and characters in an input file test1.txt.

Explanation:
This program reads text from a file and counts the number of words and characters.

C Program:

#include <stdio.h>

#include <ctype.h>

 

int main() {

    FILE *file;

    char ch;

    int words = 0, chars = 0, inWord = 0;

 

    file = fopen("test1.txt", "r");

    if (file == NULL) {

        printf("File not found!\n");

        return 1;

    }

 

    while ((ch = fgetc(file)) != EOF) {

        chars++;

        if (isspace(ch)) {

            inWord = 0;

        } else if (inWord == 0) {

            inWord = 1;

            words++;

        }

    }

 

    fclose(file);

 

    printf("Total characters: %d\n", chars);

    printf("Total words: %d\n", words);

 

    return 0;

}

Explanation of the Program:

  1. Opens test1.txt for reading.
  2. Counts characters and words.

Question 12: Write a program in C for the multiplication of two matrices of size 4x4.

Explanation:
This program multiplies two 4x4 matrices entered by the user.

C Program:

#include <stdio.h>

 

int main() {

    int a[4][4], b[4][4], result[4][4] = {0}, i, j, k;

 

    printf("Enter elements of first 4x4 matrix:\n");

    for (i = 0; i < 4; i++) {

        for (j = 0; j < 4; j++) {

            scanf("%d", &a[i][j]);

        }

    }

 

    printf("Enter elements of second 4x4 matrix:\n");

    for (i = 0; i < 4; i++) {

        for (j = 0; j < 4; j++) {

            scanf("%d", &b[i][j]);

        }

    }

 

    for (i = 0; i < 4; i++) {

        for (j = 0; j < 4; j++) {

            for (k = 0; k < 4; k++) {

                result[i][j] += a[i][k] * b[k][j];

            }

        }

    }

 

    printf("Resultant matrix:\n");

    for (i = 0; i < 4; i++) {

        for (j = 0; j < 4; j++) {

            printf("%d ", result[i][j]);

        }

        printf("\n");

    }

 

    return 0;

}

Explanation of the Program:

  1. Reads two 4x4 matrices.
  2. Multiplies them and prints the result.

Detailed Explanation for Question 8:

"Write a C program to add two complex numbers by passing structure to a function."

Concepts Involved:

  1. Complex Numbers:
    A complex number has two parts:
    • Real part
    • Imaginary part
      Example: 3+4i3 + 4i → Real part = 3, Imaginary part = 4
  2. Structures in C:
    Structures allow grouping different data types under a single name. Here, a structure can hold the real and imaginary parts of a complex number.
  3. Function with Structure as Argument:
    Passing a structure to a function allows performing operations like addition using separate logic.

Steps to Solve the Problem:

  1. Define a structure to represent a complex number with two float members: real and imaginary.
  2. Create a function that takes two complex numbers as parameters and returns their sum.
  3. Input values for two complex numbers from the user.
  4. Call the function to compute the sum.
  5. Display the result.

C Program:

#include <stdio.h>

 

// Define a structure for complex numbers

struct Complex {

    float real;

    float imag;

};

 

// Function to add two complex numbers

struct Complex addComplex(struct Complex c1, struct Complex c2) {

    struct Complex result;

    result.real = c1.real + c2.real;

    result.imag = c1.imag + c2.imag;

    return result;

}

 

int main() {

    struct Complex num1, num2, sum;

 

    // Input for first complex number

    printf("Enter real and imaginary part of first complex number:\n");

    scanf("%f %f", &num1.real, &num1.imag);

 

    // Input for second complex number

    printf("Enter real and imaginary part of second complex number:\n");

    scanf("%f %f", &num2.real, &num2.imag);

 

    // Add the two complex numbers

    sum = addComplex(num1, num2);

 

    // Display the result

    printf("Sum of the complex numbers = %.2f + %.2fi\n", sum.real, sum.imag);

 

    return 0;

}

Explanation of the Program:

  1. Structure Definition:
  2. struct Complex {
  3.     float real;
  4.     float imag;
  5. };

This structure holds the real and imaginary parts of a complex number.

  1. Function to Add Complex Numbers:
  2. struct Complex addComplex(struct Complex c1, struct Complex c2) {
  3.     struct Complex result;
  4.     result.real = c1.real + c2.real;
  5.     result.imag = c1.imag + c2.imag;
  6.     return result;
  7. }
    • Takes two Complex structures as input.
    • Adds the real parts and imaginary parts separately.
    • Returns the result as a Complex structure.
  8. User Input:
  9. scanf("%f %f", &num1.real, &num1.imag);
  10. scanf("%f %f", &num2.real, &num2.imag);

Reads real and imaginary parts for both complex numbers.

  1. Function Call and Result:
  2. sum = addComplex(num1, num2);
  3. printf("Sum of the complex numbers = %.2f + %.2fi\n", sum.real, sum.imag);

Calls the addComplex() function and prints the result.

Sample Output:

Enter real and imaginary part of first complex number:

3.5 2.5

Enter real and imaginary part of second complex number:

1.5 4.5

Sum of the complex numbers = 5.00 + 7.00i

Key Points to Remember:

  • Structures help organize related data.
  • Functions can accept structures as arguments for modularity.
  • Real and imaginary parts are added separately in complex addition.

Detailed Explanation for Question 9:

"What do you mean by pointer? Write a C program to access elements in a 2-dimensional array using a pointer."

Concepts Involved:

1. Pointer in C:

A pointer is a variable that stores the memory address of another variable. It is used to access and manipulate data stored in memory directly.

  • Declaration:
  • int *ptr;  // Pointer to an integer
  • Usage:
    • & (Address-of operator): Gives the address of a variable.
    • * (Dereference operator): Accesses the value at the address stored in a pointer.

2. 2D Array in C:

A 2D array is an array of arrays.
Example:

int arr[2][3] = { {1, 2, 3}, {4, 5, 6} };

Memory Representation:

Row 0 → 1 2 3 

Row 1 → 4 5 6

3. Pointer with 2D Arrays:

In C, a 2D array is stored in row-major order, meaning the elements are stored row by row in memory.

  • Accessing elements:
    Using pointers, an element at arr[i][j] can be accessed as:
  • *(*(arr + i) + j)

Program to Access 2D Array Elements Using Pointers

#include <stdio.h>

 

#define ROW 2

#define COL 3

 

int main() {

    int arr[ROW][COL] = { {1, 2, 3}, {4, 5, 6} };

    int *ptr;

   

    // Pointer pointing to the first element of the array

    ptr = &arr[0][0];

 

    printf("Accessing 2D array elements using pointer:\n");

 

    // Loop through the 2D array using pointer arithmetic

    for(int i = 0; i < ROW; i++) {

        for(int j = 0; j < COL; j++) {

            // Using pointer to access elements

            printf("%d ", *(ptr + i * COL + j));

        }

        printf("\n");

    }

 

    return 0;

}

Explanation of the Program:

  1. Array Initialization:
  2. int arr[ROW][COL] = { {1, 2, 3}, {4, 5, 6} };

Creates a 2D array with 2 rows and 3 columns.

  1. Pointer Declaration and Initialization:
  2. int *ptr;
  3. ptr = &arr[0][0];

The pointer ptr is assigned the address of the first element of the array.

  1. Accessing Elements Using Pointer Arithmetic:
  2. *(ptr + i * COL + j)
    • ptr + i * COL + j moves the pointer to the desired element.
    • * dereferences the pointer to access the value.
  3. Nested Loops for Traversing:
    Loops iterate over each row and column to print the array.

Memory Representation:

In memory, the 2D array is stored as:

Index: 0 1 2 3 4 5 

Value: 1 2 3 4 5 6

So,

  • *(ptr + 0) → 1
  • *(ptr + 1) → 2
  • *(ptr + 3) → 4

Sample Output:

Accessing 2D array elements using pointer:

1 2 3

4 5 6

Alternate Method Using Double Pointers:

#include <stdio.h>

 

#define ROW 2

#define COL 3

 

void printArray(int (*ptr)[COL]) {

    for(int i = 0; i < ROW; i++) {

        for(int j = 0; j < COL; j++) {

            printf("%d ", *(*(ptr + i) + j));

        }

        printf("\n");

    }

}

 

int main() {

    int arr[ROW][COL] = { {1, 2, 3}, {4, 5, 6} };

    printArray(arr);

    return 0;

}

Explanation:

  • int (*ptr)[COL] is a pointer to an array of integers with COL elements.
  • *(*(ptr + i) + j) accesses elements.

Key Points to Remember:

  • A pointer can access all elements of a 2D array through pointer arithmetic.
  • The formula *(ptr + i * COL + j) is based on row-major order.
  • Efficient in terms of memory and performance, especially for large arrays.

Detailed Explanation for Question 10:

"How is a string different from an array? Write a C program to convert a lower case string to upper case string without using string manipulation functions."

Part 1: Difference Between a String and an Array

Aspect

Array

String

Definition

Collection of elements of the same data type.

Array of characters ending with a null character (\0).

Data Types

Can be of any type (int, float, char, etc.).

Always an array of characters.

Null Terminator

Does not have a null character automatically.

Ends with \0 to mark the end of the string.

Access

Accessed by index.

Accessed by index but ends at \0.

Example

int arr[5] = {1, 2, 3, 4, 5};

char str[] = "hello"; → { 'h', 'e', 'l', 'l', 'o', '\0' }

Operations

No direct functions for operations.

Has built-in functions (strlen(), strcpy()).

Part 2: C Program to Convert Lowercase to Uppercase Without String Functions

Concept Used:

  • ASCII Values:
    • Lowercase letters (a–z) have ASCII values from 97 to 122.
    • Uppercase letters (A–Z) have ASCII values from 65 to 90.
    • Difference: A - a = -32 → To convert lowercase to uppercase, subtract 32.

Logic:

  • Traverse each character in the string.
  • If the character is between 'a' and 'z', convert it by subtracting 32.
  • Else, leave it unchanged.

C Program:

#include <stdio.h>

 

void toUpperCase(char *str) {

    int i = 0;

   

    // Traverse each character in the string

    while (str[i] != '\0') {

        // Check if the character is lowercase

        if (str[i] >= 'a' && str[i] <= 'z') {

            str[i] = str[i] - 32;  // Convert to uppercase

        }

        i++;

    }

}

 

int main() {

    char str[100];

 

    // Input from the user

    printf("Enter a lowercase string: ");

    scanf("%[^\n]%*c", str);  // Reads string with spaces

 

    // Convert to uppercase

    toUpperCase(str);

 

    // Display result

    printf("Uppercase string: %s\n", str);

 

    return 0;

}

Explanation of the Program:

  1. Function to Convert to Uppercase:
  2. void toUpperCase(char *str)
    • Takes a character array (string) as input.
    • Loops through the string until '\0' is reached.
  3. Conversion Logic:
  4. if (str[i] >= 'a' && str[i] <= 'z') {
  5.     str[i] = str[i] - 32;
  6. }
    • If the character is lowercase, subtract 32 to convert it to uppercase.
  7. User Input:
  8. scanf("%[^\n]%*c", str);
    • Reads the entire input, including spaces.
  9. Output:
  10. printf("Uppercase string: %s\n", str);
    • Prints the converted uppercase string.

Sample Run:

Input:

Enter a lowercase string: hello world

Output:

Uppercase string: HELLO WORLD

Key Points:

  • ASCII conversion avoids using built-in functions like strupr() or toupper().
  • The program preserves non-alphabetic characters (spaces, punctuation).
  • Efficient and lightweight solution.

Would you like me to explain any part of this further or modify the program?

Detailed Explanation for Question 11:

"Write a program in C to count the number of words and characters in an input file test1.txt."

Concepts Involved:

  1. File Handling in C:
    • C uses file pointers (FILE *) to handle files.
    • Functions like fopen(), fgetc(), and fclose() are used to read/write files.
  2. Character Counting:
    • Count every character until the End of File (EOF) is reached.
  3. Word Counting:
    • A word is detected when a sequence of characters is followed by a space, newline, or tab.
    • Handle multiple spaces or line breaks to avoid miscounts.

C Program to Count Words and Characters in a File

#include <stdio.h>

#include <stdlib.h>

#include <ctype.h>

 

int main() {

    FILE *file;

    char filename[] = "test1.txt";

    char ch;

    int characters = 0, words = 0;

    int inWord = 0;

 

    // Open the file in read mode

    file = fopen(filename, "r");

 

    // Check if the file was opened successfully

    if (file == NULL) {

        printf("Error: Could not open the file %s\n", filename);

        return 1;

    }

 

    // Read each character until EOF

    while ((ch = fgetc(file)) != EOF) {

        characters++;  // Count every character

 

        // Check for word boundaries

        if (isspace(ch)) {

            inWord = 0;  // Outside a word when whitespace is found

        } else if (inWord == 0) {

            inWord = 1;  // Start of a new word

            words++;

        }

    }

 

    // Close the file

    fclose(file);

 

    // Display the counts

    printf("Total Characters: %d\n", characters);

    printf("Total Words: %d\n", words);

 

    return 0;

}

Explanation of the Program:

  1. File Pointer Declaration:
  2. FILE *file;
    • Declares a file pointer to manage file operations.
  3. Opening the File:
  4. file = fopen(filename, "r");
    • Opens test1.txt in read mode ("r").
    • If the file doesn't exist, it prints an error.
  5. Reading the File:
  6. while ((ch = fgetc(file)) != EOF)
    • Reads each character one by one using fgetc() until End of File (EOF).
  7. Character Counting:
  8. characters++;
    • Increments for every character read, including spaces and newlines.
  9. Word Counting Logic:
  10. if (isspace(ch)) {
  11.     inWord = 0;
  12. } else if (inWord == 0) {
  13.     inWord = 1;
  14.     words++;
  15. }
    • isspace(ch) checks for spaces, tabs, or newlines.
    • inWord tracks if the current character is inside a word.
  16. Closing the File:
  17. fclose(file);
    • Always close the file to free system resources.
  18. Displaying the Result:
  19. printf("Total Characters: %d\n", characters);
  20. printf("Total Words: %d\n", words);

Sample Content of test1.txt:

Hello World!

This is a file with words.

Program Output:

Total Characters: 35 

Total Words: 7

Explanation:

  • Characters (35): Includes all letters, spaces, punctuation, and newline characters.
  • Words (7): Hello, World!, This, is, a, file, with, words.

Edge Cases to Consider:

  1. Empty File:
    • Output → Characters: 0, Words: 0
  2. Multiple Spaces or Newlines:
    • Handled correctly due to the use of isspace().
  3. Special Characters:
    • Counted as characters but not as words.

Key Points:

  • File handling ensures reading the file content efficiently.
  • Word detection is based on whitespace separation.
  • Proper error handling prevents crashes if the file is missing.

Detailed Explanation for Question 12:

"Write a program in C for the multiplication of two matrices of size 4×4."

Concepts Involved:

1. Matrix Multiplication Basics:

Matrix multiplication involves multiplying rows of the first matrix by columns of the second matrix.

  • Formula:
    For two matrices A and B, the element at position (i, j) in the result matrix C is calculated as: C[i][j]=∑k=0n−1A[i][k]×B[k][j]C[i][j] = \sum_{k=0}^{n-1} A[i][k] \times B[k][j] Where:
    • i → Row index of matrix A.
    • j → Column index of matrix B.
    • k → Iterates over the shared dimension.

2. Matrix Dimension Compatibility:

  • Matrix A of size m x n can only be multiplied with Matrix B of size n x p.
  • The resulting matrix will be of size m x p.
  • In this case: Both matrices are 4×4, so the result will also be 4×4.

C Program for 4×4 Matrix Multiplication

#include <stdio.h>

 

#define SIZE 4  // Define the size of the matrix

 

void multiplyMatrices(int A[SIZE][SIZE], int B[SIZE][SIZE], int result[SIZE][SIZE]) {

    // Initialize the result matrix to zero

    for (int i = 0; i < SIZE; i++) {

        for (int j = 0; j < SIZE; j++) {

            result[i][j] = 0;

        }

    }

 

    // Perform matrix multiplication

    for (int i = 0; i < SIZE; i++) {

        for (int j = 0; j < SIZE; j++) {

            for (int k = 0; k < SIZE; k++) {

                result[i][j] += A[i][k] * B[k][j];

            }

        }

    }

}

 

// Function to print the matrix

void printMatrix(int matrix[SIZE][SIZE]) {

    for (int i = 0; i < SIZE; i++) {

        for (int j = 0; j < SIZE; j++) {

            printf("%d\t", matrix[i][j]);

        }

        printf("\n");

    }

}

 

int main() {

    // Define two 4x4 matrices

    int A[SIZE][SIZE] = {

        {1, 2, 3, 4},

        {5, 6, 7, 8},

        {9, 10, 11, 12},

        {13, 14, 15, 16}

    };

 

    int B[SIZE][SIZE] = {

        {16, 15, 14, 13},

        {12, 11, 10, 9},

        {8, 7, 6, 5},

        {4, 3, 2, 1}

    };

 

    int result[SIZE][SIZE];  // Result matrix

 

    // Multiply matrices A and B

    multiplyMatrices(A, B, result);

 

    // Display the result

    printf("Resultant Matrix after Multiplication:\n");

    printMatrix(result);

 

    return 0;

}

Explanation of the Program:

  1. Matrix Initialization:
    Two 4×4 matrices A and B are defined with predefined values.
  2. Matrix Multiplication Logic:
  3. result[i][j] += A[i][k] * B[k][j];
    • Three nested loops are used:
      • Outer loop (i): Iterates over rows of matrix A.
      • Middle loop (j): Iterates over columns of matrix B.
      • Inner loop (k): Multiplies and sums the corresponding elements.
  4. Result Matrix Initialization:
    The result matrix is initialized to zero before starting the multiplication.
  5. Matrix Printing:
    A separate function printMatrix() is used to display the resulting matrix in a formatted way.

Step-by-Step Example Calculation:

Given:

Matrix A:

1  2  3  4 

5  6  7  8 

9 10 11 12 

13 14 15 16

Matrix B:

16 15 14 13 

12 11 10  9 

8  7  6  5 

4  3  2  1

Calculation for Element at (0,0):

C[0][0]=(1×16)+(2×12)+(3×8)+(4×4)=16+24+24+16=80C[0][0] = (1 \times 16) + (2 \times 12) + (3 \times 8) + (4 \times 4) = 16 + 24 + 24 + 16 = 80

Sample Output:

Resultant Matrix after Multiplication:

80           70           60           50          

240        214        188        162       

400        358        316        274       

560        502        444        386

Key Points to Remember:

  • Time Complexity: O(n3)O(n^3) due to three nested loops.
  • Space Complexity: Requires additional space for the result matrix.
  • Matrix Compatibility: Both matrices must have the same size for this program.

Detailed Explanation for Question 13:

"Create a structure named 'Book' to store book details like title, author, and price. Write a C program to input details for three books, find the most expensive and the lowest priced books, and display their information."

Concepts Involved:

  1. Structures in C:
    • Used to group related data of different types under a single name.
    • Allows combining variables like strings (for title and author) and floats (for price).
  2. Array of Structures:
    • An array can store multiple instances of a structure, making it easy to handle multiple books.
  3. Input/Output Operations:
    • Using scanf() and printf() to take user input and display output.
  4. Finding Maximum and Minimum Values:
    • Comparing prices to determine the most expensive and cheapest books.

C Program to Handle Book Details

#include <stdio.h>

#include <string.h>

 

// Define a structure for Book

struct Book {

    char title[100];

    char author[100];

    float price;

};

 

int main() {

    struct Book books[3];  // Array to store 3 books

    int i, maxIndex = 0, minIndex = 0;

 

    // Input details for 3 books

    for (i = 0; i < 3; i++) {

        printf("Enter details for Book %d:\n", i + 1);

       

        // Input title

        printf("Title: ");

        getchar();  // Clears the buffer

        fgets(books[i].title, sizeof(books[i].title), stdin);

        books[i].title[strcspn(books[i].title, "\n")] = '\0';  // Remove newline character

 

        // Input author

        printf("Author: ");

        fgets(books[i].author, sizeof(books[i].author), stdin);

        books[i].author[strcspn(books[i].author, "\n")] = '\0';  // Remove newline character

 

        // Input price

        printf("Price: ");

        scanf("%f", &books[i].price);

    }

 

    // Find the most expensive and the cheapest books

    for (i = 1; i < 3; i++) {

        if (books[i].price > books[maxIndex].price) {

            maxIndex = i;  // Most expensive book

        }

        if (books[i].price < books[minIndex].price) {

            minIndex = i;  // Cheapest book

        }

    }

 

    // Display the most expensive book

    printf("\nMost Expensive Book:\n");

    printf("Title: %s\n", books[maxIndex].title);

    printf("Author: %s\n", books[maxIndex].author);

    printf("Price: %.2f\n", books[maxIndex].price);

 

    // Display the cheapest book

    printf("\nCheapest Book:\n");

    printf("Title: %s\n", books[minIndex].title);

    printf("Author: %s\n", books[minIndex].author);

    printf("Price: %.2f\n", books[minIndex].price);

 

    return 0;

}

Explanation of the Program:

  1. Structure Definition:
  2. struct Book {
  3.     char title[100];
  4.     char author[100];
  5.     float price;
  6. };
    • title → Stores the book title.
    • author → Stores the author's name.
    • price → Stores the price of the book.
  7. Array of Structures:
  8. struct Book books[3];
    • Stores details of 3 books.
  9. Input for Book Details:
  10. fgets(books[i].title, sizeof(books[i].title), stdin);
  11. fgets(books[i].author, sizeof(books[i].author), stdin);
  12. scanf("%f", &books[i].price);
    • fgets() is used for multi-word input.
    • scanf() reads the price.
    • getchar() clears the input buffer to avoid skipping input.
  13. Finding Maximum and Minimum Prices:
  14. if (books[i].price > books[maxIndex].price)
  15.     maxIndex = i;
  16. if (books[i].price < books[minIndex].price)
  17.     minIndex = i;
    • Iterates through the books and tracks the index of the most expensive (maxIndex) and the cheapest (minIndex) books.
  18. Displaying Results:
  19. printf("\nMost Expensive Book:\n");
  20. printf("Title: %s\n", books[maxIndex].title);
  21. printf("Author: %s\n", books[maxIndex].author);
  22. printf("Price: %.2f\n", books[maxIndex].price);
    • Prints the details of the most expensive and the cheapest books.

Sample Run:

Input:

Enter details for Book 1:

Title: The Great Gatsby

Author: F. Scott Fitzgerald

Price: 450

 

Enter details for Book 2:

Title: 1984

Author: George Orwell

Price: 300

 

Enter details for Book 3:

Title: War and Peace

Author: Leo Tolstoy

Price: 600

Output:

Most Expensive Book:

Title: War and Peace

Author: Leo Tolstoy

Price: 600.00

 

Cheapest Book:

Title: 1984

Author: George Orwell

Price: 300.00

Key Points:

  • Structures allow grouping related data types, making the program clean and organized.
  • Array of structures efficiently stores multiple records.
  • String handling with fgets() and strcspn() ensures correct multi-word input.
  • Comparison logic correctly identifies the most expensive and cheapest books.

 

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