Swap Two Numbers in C

Algorithm

1. Read a and b.
2. Call swap with the addresses of a and b.
3. Inside swap, copy *x to a temporary, copy *y into *x, then copy the temporary into *y.
4. Print the values after the swap.

C program

#include <stdio.h>

void swap(int *x, int *y) {
    int temp = *x;
    *x = *y;
    *y = temp;
}

int main(void) {
    int a, b;

    printf("Enter two integers:\n");
    if (scanf("%d %d", &a, &b) != 2) {
        printf("Error: invalid input\n");
        return 1;
    }

    swap(&a, &b);
    printf("After swap: a = %d, b = %d\n", a, b);
    return 0;
}

Sample input and output

3 8

Enter two integers:
After swap: a = 8, b = 3

4 4

Enter two integers:
After swap: a = 4, b = 4

How the program works

A function receives copies of its arguments, so swapping two int parameters would change only the copies. Passing the addresses lets swap change a and b in main.

The temporary variable is needed because the first assignment overwrites one of the values. Without it both variables would end up holding the same number.

Viva questions

Common mistakes

• Passing a and b by value, so main never sees the swap.
• Forgetting the * in the function body, which writes to the pointer itself rather than the value.