Question

I’m very confused why this code works. I was taught that a pointer was a “capsule” holding the address of something. So the swapnum function should be expecting one of those “capsules”, not the actual address of the int. Is it creating a temporary pointer and setting it to that address? If that’s the case, then what would be happening if I passed a pointer in by reference such as int * c = &a; swapnum (&c...?

#include <stdio.h>

void swapnum(int *i, int *j) {
  int temp = *i;
  *i = *j;
  *j = temp;
}

int main(void) {
  int a = 10;
  int b = 20;

  swapnum(&a, &b);
  printf("A is %d and B is %d\n", a, b);
  return 0;
}

Answer 1

pointer is just another word for address. There’s nothing really magic about it, it is just a variable that contains a number, that number being the memory address of the thing it points to.

In your example, memory might look like this:

0x00001004 - 10         # a
0x00001008 - 20         # b
0x0000100C - 0x0001004  # i
0x00001010 - 0x0001008  # j
0x00001014 - 00         # temp

i and j are both int* which means they contain an address that contains an integar. The code *i says “return whatever value is at the address in i and assume it is an int.

So if you look at your swapnum, here’s what happens:

int temp = *i;
# take the value at 0x0001004 (10) and place it in 0x00001014

*i = *j;
# place the value at 0x0001008 (20) and place it in 0x00001004

*j = temp;
# place the value at 0x0001014 (10) and place it in 0x0001008

After which memory looks like this:

0x00001004 - 20         # a
0x00001008 - 10         # b
0x0000100C - 0x0001004  # i
0x00001010 - 0x0001008  # j
0x00001014 - 10         # temp

You could certainly do this:

int* c = &a;
swapnum(c, &b);

All you are doing is getting the address of a, putting it in c and then passing it into swapnum, which expects an address. A pointer is just an address.