Question

I just write a simple program to print the address of variable.

#include <stdio.h>

void main(){
  int *a;
  int b;

  b=5;
  a=&b;

  printf("\n*a is the value of the integer:  %d\n",*a);
  printf("&a is the address of variable a: %p\n", &a);
  printf(" a is the address stored in a:   %p\n",a);
  printf("&b is the address of variable b: %p\n\n",&b);
}

The result is:

*a is the value of the integer:  5
&a is the address of variable a: 0x7fff935ad2b0
 a is the address stored in a:   0x7fff935ad2bc
&b is the address of variable b: 0x7fff935ad2bc

Why the addresses printed are not in the standard form?

Sorry for the silly question, but I just don’t understand.

EDIT

I have another program

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

struct small{
  int a;
};

int main(){
   struct small *sm;

   sm = malloc(sizeof(struct small));
   memset(sm,0,sizeof(struct small));

   sm->a = 5;

   printf("The address of sm is: %p\n", &sm);
   printf("The address stored in sm is: %p\n", sm);

   return 0;
}

The output is:

The address of sm is: 0x7fffd1363158
The address stored in sm is: 0x17a3010

So I expected the format 0x17a3010 is somewhat standard to me.

And again why in this case the format of %p is different?

Answer 1

That is called hexadecimal, i.e., base-16. It makes it easier to view groups of bits as bytes than binary or base-10 would. I’m not sure I understand your question. There is no “standard form”, but if there were one this would be it.

Per your edit:

The output is:

The address of sm is: 0x7fffd1363158
The address stored in sm is: 0x17a3010
So I expected the format 0x17a3010 is somewhat standard to me.

And again why in this case the format of %p is different?

Those two are exactly the same. 0x17a3010 is no different than 0x00000000017a3010.