Written: August 14, 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.
Common mistakes
Forgetting a semicolon, using the wrong format specifier in printf/scanf, and mixing up assignment (=) with comparison (==) are the usual culprits.
Compile with warnings enabled (`gcc -Wall`) so the compiler points at risky casts and unused variables before you chase them by hand.
Try this next
Change the sample inputs, add a second test case, and briefly note what stayed the same. Teaching yourself with tiny experiments sticks better than copying a longer program you never run.
When you can explain every line without looking, you are ready for the next exercise in the series.
