Pointers & Functions

C is always pass-by-value

Function arguments are always copied. To let a function change a caller's variable, pass a pointer to it instead of the value itself — this is how C simulates "pass by reference".

swap with pointers

Passing &a and &b lets swap reach into the caller's memory directly.

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

int main(void) {
    int a = 1, b = 2;
    swap(&a, &b);
    printf("%d %d\n", a, b);
    return 0;
}

Output:

2 1

Quick check: Why does `swap(int x, int y)` (no pointers) fail to swap the caller's variables?

• C doesn't allow swapping
• The compiler optimizes it away
• Pointers cannot be passed
• x and y are copies of the originals