# Pointers & Functions — C Programming

Source: https://www.geekswithgeeks.com/en/c/c-pointers-functions

> Pass a variable's address so a function can modify the original.

## 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.

```c
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
```

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

- [x] x and y are copies of the originals
- [ ] C doesn't allow swapping
- [ ] The compiler optimizes it away

*Answer:* x and y are copies of the originals. Without pointers, the function only swaps its own local copies.