Subtracting the values of two pointers to an object results in a signed integer *1. So it would be printed using at least the d conversion specifier.

To make sure there is a type being wide enough to hold such a “pointer-difference”, since C99 <stddef.h> defines the type ptrdiff_t. To print a ptrdiff_t use the t length modifier.

#include <stdlib.h> /* for EXIT_SUCCESS */
#include <stdio.h> /* for printf() */
#include <stddef.h> /* for ptrdiff_t */

int main(void)
{
  int a[2];
  int * p1 = &a[0], * p2 = &a[1];
  ptrdiff_t pd = p2 - p1;

  printf("p1 = %p\\n", (void*) p1);
  printf("p2 = %p\\n", (void*) p2);
  printf("p2 - p1 = %td\\n", pd);

  return EXIT_SUCCESS;
}

The result might look like this:

p1 = 0x7fff6679f430
p2 = 0x7fff6679f434
p2 - p1 = 1

Please note that the resulting value of the difference is scaled by the size of the type the pointers subtracted point to, an int here. The size of an int for this example is 4.

*1If the two pointers to be subtracted do not point to the same object the behaviour is undefined.