Regarding pointer arithmetic in C

Question

I'm trying to print the address a pointer is holding in C. Would I use %d or %p for the format specifier? And are all pointer addresses ints? In this youtube video, I thought that was the case but when I tried to code something, it would not compile unless I used %p. Also, isn't i_ptr+1 supposed to be the memory address i_ptr holds + 4 because an int is 4 bytes? https://www.youtube.com/watch?v=JTttg85xsbo

Thankyou!!!!

#include <stdio.h>

void test(int *i_ptr);
int main() {
  int i[5];
  test(i);
  return 0;
}

void test(int *i_ptr) {
  printf("%d\n", i_ptr);
  printf("%d\n", i_ptr + 1);
}

~

Answer 1

If you have a compiler that supports C99, you can use the type uintptr_t to hold the value of a pointer. However, since its size is not specified by compiler and there isn't a printf specifier for such a type, you have to experiment with using "%d", "%ld" and "%lld" to make things work. I want to emphasize that it is not portable code.

The following code worked for me, using gcc 4.8.2 on a 64-bit Linux machine with -std=c99.

#include <stdio.h>
#include <stdint.h>

void test(int *i_ptr);

int main() {
   int i[5];
   test(i);
   return 0;
}

void test(int *i_ptr) {
   uintptr_t p = (uintptr_t)i_ptr;
   printf("%ld\n", p);
   p = (uintptr_t)(i_ptr+1);
   printf("%ld\n", p);
}

With the following output from a sample run.

140735074589984
140735074589988

Update

The format specifier "%td" is meant to be used for ptrdiff_t. It's more likely that it will work for uintptr_t as well.