Exercise 1: Finding Minimum and Maximum Values in Arrays

Implementation 1: Using Pointer Parameters

#include <stdio.h>
#define N 5

void input_data(int arr[], int n);
void display_data(int arr[], int n);
void find_min_max(int arr[], int n, int *pmin, int *pmax);

int main() {
    int numbers[N];
    int minimum, maximum;
    
    printf("Enter %d integers:\n", N);
    input_data(numbers, N);
    
    printf("Original data: \n");
    display_data(numbers, N);
    
    printf("Processing data...\n");
    find_min_max(numbers, N, &minimum, &maximum);
    
    printf("Results:\n");
    printf("min = %d, max = %d\n", minimum, maximum);
    
    return 0;
}

void input_data(int arr[], int n) {
    int i;
    for (i = 0; i < n; ++i)
        scanf("%d", &arr[i]);
}

void display_data(int arr[], int n) {
    int i;
    for (i = 0; i < n; ++i)
        printf("%d ", arr[i]);
    printf("\n");
}

void find_min_max(int arr[], int n, int *pmin, int *pmax) {
    int i;
    *pmin = *pmax = arr[0];
    
    for (i = 0; i < n; ++i) {
        if (arr[i] < *pmin)
            *pmin = arr[i];
        else if (arr[i] > *pmax)
            *pmax = arr[i];
    }
}

Analysis:

• The function finds both minimum and maximum values from the input array
• Both pointers initially point to the first element of the array

Implementation 2: Returning Pointer to Maximum

#include <stdio.h>
#define N 5

void input_data(int arr[], int n);
void display_data(int arr[], int n);
int *locate_maximum(int arr[], int n);

int main() {
    int numbers[N];
    int *pmax;
    
    printf("Enter %d integers:\n", N);
    input_data(numbers, N);
    
    printf("Original data: \n");
    display_data(numbers, N);
    
    printf("Processing data...\n");
    pmax = locate_maximum(numbers, N);
    
    printf("Results:\n");
    printf("max = %d\n", *pmax);
    
    return 0;
}

void input_data(int arr[], int n) {
    int i;
    for (i = 0; i < n; ++i)
        scanf("%d", &arr[i]);
}

void display_data(int arr[], int n) {
    int i;
    for (i = 0; i < n; ++i)
        printf("%d ", arr[i]);
    printf("\n");
}

int *locate_maximum(int arr[], int n) {
    int max_index = 0;
    int i;
    
    for (i = 0; i < n; ++i)
        if (arr[i] > arr[max_index])
            max_index = i;
    
    return &arr[max_index];
}

Analysis:

• This function returns the address of the maximum element in the array
• The return value is a pointer to an integer

Exercise 2: String Operations with Arrays vs Pointers

Implementation 1: Character Arrays

#include <stdio.h>
#include <string.h>
#define N 80

int main() {
    char s1[N] = "Learning makes me happy";
    char s2[N] = "Learning makes me sleepy";
    char buffer[N];
    
    printf("sizeof(s1) vs. strlen(s1): \n");
    printf("sizeof(s1) = %d\n", sizeof(s1));
    printf("strlen(s1) = %d\n", strlen(s1));
    
    printf("\nbefore swap: \n");
    printf("s1: %s\n", s1);
    printf("s2: %s\n", s2);
    
    printf("\nswapping...\n");
    strcpy(buffer, s1);
    strcpy(s1, s2);
    strcpy(s2, buffer);
    
    printf("\nafter swap: \n");
    printf("s1: %s\n", s1);
    printf("s2: %s\n", s2);
    
    return 0;
}

Analysis:

• sizeof(s1) returns 80 (total allocated space for the array)
• strlen(s1) returns 23 (actual string length without null terminator)
• Direct asssignment like s1 = "string"; is illegal for character arrays
• The swap operation works because strcpy copies actual contant

Implementation 2: Character Pointers

#include <stdio.h>
#include <string.h>
#define N 80

int main() {
    char *s1 = "Learning makes me happy";
    char *s2 = "Learning makes me sleepy";
    char *tmp;
    
    printf("sizeof(s1) vs. strlen(s1): \n");
    printf("sizeof(s1) = %d\n", sizeof(s1));
    printf("strlen(s1) = %d\n", strlen(s1));
    
    printf("\nbefore swap: \n");
    printf("s1: %s\n", s1);
    printf("s2: %s\n", s2);
    
    printf("\nswapping...\n");
    tmp = s1;
    s1 = s2;
    s2 = tmp;
    
    printf("\nafter swap: \n");
    printf("s1: %s\n", s1);
    printf("s2: %s\n", s2);
    
    return 0;
}

Analysis:

• sizeof(s1) returns 4 or 8 (size of pointer variable on the system)
• strlen(s1) returns 23 (length of the string literal)
• No issues with this approach
• The swap exchanges pointer values, not the string data itself

Exercise 3: Two-Dimensional Arrays and Pointers

#include <stdio.h>


int main() {
    int matrix[2][4] = {{1, 9, 8, 4}, {2, 0, 4, 9}};
    int i, j;
    int *ptr_element;
    int (*ptr_row)[4];
    
    printf("Output 1: Using array name with indices\n");
    for (i = 0; i < 2; ++i) {
        for (j = 0; j < 4; ++j)
            printf("%d ", matrix[i][j]);
        printf("\n");
    }
    
    printf("\nOutput 2: Using element pointer\n");
    for (ptr_element = &matrix[0][0], i = 0; ptr_element < &matrix[0][0] + 8; ++ptr_element, ++i) {
        printf("%d ", *ptr_element);
        if ((i + 1) % 4 == 0)
            printf("\n");
    }
    
    printf("\nOutput 3: Using row pointer\n");
    for (ptr_row = matrix; ptr_row < matrix + 2; ++ptr_row) {
        for (j = 0; j < 4; ++j)
            printf("%d ", *(*ptr_row + j));
        printf("\n");
    }
    
    return 0;
}

Analysis:

• ptr_element is a pointer to a single integer element
• ptr_row is a pointer to an array of 4 integers (array pointer)

Exercise 4: Character Replacement in Strings

#include <stdio.h>
#define N 80

void substitute(char *str, char old_char, char new_char);

int main() {
    char text[N] = "Programming is difficult or not, it is a question.";
    
    printf("Original text: \n");
    printf("%s\n", text);
    
    substitute(text, 'i', '*');
    
    printf("Modified text: \n");
    printf("%s\n", text);
    
    return 0;
}

void substitute(char *str, char old_char, char new_char) {
    while (*str) {
        if (*str == old_char)
            *str = new_char;
        str++;
    }
}

Analysis:

• The function replaces all occurrences of old_char with new_char
• Works correctly with pointer arithmetic

Exercise 5: String Truncation

#include <stdio.h>
#define N 80

char* truncate_string(char* str, char delimiter);

int main() {
    char str[N];
    char ch;
    
    while (printf("Input string: "), gets(str) != NULL) {
        printf("Enter delimiter character: ");
        ch = getchar();
        
        printf("Truncating...\n");
        truncate_string(str, ch);
        printf("Truncated string: %s\n\n", str);
        
        getchar();
    }
    
    return 0;
}

char* truncate_string(char* str, char delimiter) {
    while (*str) {
        if (*str == delimiter) {
            *str = '\0';
            break;
        }
        str++;
    }
    return str;
}

-analysis:

• The function truncates the string at the first occurrence of the delimiter
• The leftover newline from getchar() needs to be consumed with an additional getchar() call

Exercise 6: ID Number Validation

#include <stdio.h>
#include <string.h>
#define N 5

int validate_id(char* str);

int main() {
    char* id_list[N] = {
        "31010120000721656X",
        "3301061996X0203301",
        "53010220051126571",
        "510104199211197977",
        "53010220051126133Y"
    };
    int i;
    
    for (i = 0; i < N; ++i)
        if (validate_id(id_list[i]))
            printf("%s\tTrue\n", id_list[i]);
        else
            printf("%s\tFalse\n", id_list[i]);
    
    return 0;
}

int validate_id(char* str) {
    int position = 0;
    
    while (*str) {
        position++;
        if (*str < '0' || *str > '9') {
            if (*str != 'X' || position != 18) {
                return 0;
            }
        }
        str++;
    }
    
    if (position != 18) {
        return 0;
    }
    return 1;
}

Enalysis:

• Validates Chinese ID numbers (should be 18 characters)
• Allows digits 0-9, and only allows 'X' at position 18

Exercise 7: Caesar Cipher Encoding and Decoding

#include <stdio.h>
#define N 80

void encode(char* str, int shift);
void decode(char* str, int shift);

int main() {
    char text[N];
    int n;
    
    printf("Enter English text: ");
    gets(text);
    
    printf("Enter shift value: ");
    scanf("%d", &n);
    
    printf("Encoded text: ");
    encode(text, n);
    printf("%s\n", text);
    
    printf("Decoded text: ");
    decode(text, n);
    printf("%s\n", text);
    
    return 0;
}

void encode(char* str, int shift) {
    while (*str) {
        if (*str >= 'A' && *str <= 'Z') {
            *str = (*str - 'A' + shift % 26 + 26) % 26 + 'A';
        }
        else if (*str >= 'a' && *str <= 'z') {
            *str = (*str - 'a' + shift % 26 + 26) % 26 + 'a';
        }
        str++;
    }
}

void decode(char* str, int shift) {
    while (*str) {
        if (*str >= 'A' && *str <= 'Z') {
            *str = (*str - 'A' + 26 - shift % 26) % 26 + 'A';
        }
        else if (*str >= 'a' && *str <= 'z') {
            *str = (*str - 'a' + 26 - shift % 26) % 26 + 'a';
        }
        str++;
    }
}

Analysis:

• Implements Caesar cipher for letter substitution
• Handles both uppercase and lowercase letters
• Wraps around the alphabet using modulo arithmetic

Exercise 8: Command Line Arguments

Implementation 1: Simple Echo

#include <stdio.h>

int main(int argc, char* argv[]) {
    int i;
    
    for (i = 1; i < argc; ++i)
        printf("hello, %s\n", argv[i]);
    
    return 0;
}

Implementation 2: Sorting Arguments

#include <stdio.h>
#include <string.h>


int main(int argc, char* argv[]) {
    int swapped;
    char* temp;
    int i, j;
    
    for (i = 0; i < argc - 1; i++) {
        swapped = 1;
        for (j = 0; j < argc - i - 1; j++) {
            if (strcmp(argv[j], argv[j + 1]) > 0) {
                temp = argv[j];
                argv[j] = argv[j + 1];
                argv[j + 1] = temp;
            }
            swapped = 0;
        }
        if (swapped) break;
    }
    
    for (i = 1; i < argc; ++i)
        printf("hello, %s\n", argv[i]);
    
    return 0;
}

Analysis:

• argc contains the number of command line arguments
• argv[0] is the program name, argv[1] onwards are the actual arguments
• Implements bubble sort to order the arguments alphabetically