Written: August 14, 2026 · Updated: August 15, 2026
Pointer swaps show why C passes addresses when a function must change the caller’s variables.
This walkthrough keeps the logic small and readable so you can type it, run it, and then change one input at a time to see what happens.
What you will build
A swap function receives int pointers, uses a temporary int, and writes through the pointers so both originals change.
- Write void swap(int x, int y)
- Use a temp to exchange x and y
- In main, call swap(&a, &b)
- Print before and after
Working C example
include <stdio.h>
void swap(int x, int y) {
int temp = x;
x = y;
y = temp;
}
int main(void) {
int a = 10, b = 20;
printf("Before: %d %d\n", a, b);
swap(&a, &b);
printf("After: %d %d\n", a, b);
return 0;
}How the logic works
Read the program top to bottom: includes and main first, then the statements that change variables, then the print that proves the result.
If your output looks wrong, print the variables before and after the critical lines. That single habit catches most beginner bugs faster than rewriting the whole file.
If you pass a and b without &, the function only swaps copies and main stays unchanged.
Keep learning
If this walkthrough on C Program to Swap Values Using Pointers helped, open the code again and change one input or assumption. Small experiments beat rereading the same example.
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