Aren't address supposed to work like normal numbers (int)?

Question

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


int main(){
    int i[5] = {0};
    printf("%d - %d = %d" , i+1 , i , (i+1)-(i));
    return 0;
}

I know that an int is 4 bytes but aren't address supposed to work like normal numbers (int)?

Please learn about expressions. Expressions have types, too! — wildplasser, Jul 02 '20 at 00:16
@Saad Jlil: [No.](https://en.cppreference.com/w/c/language/operator_arithmetic#Pointer_arithmetic) — CherryDT, Jul 02 '20 at 00:17

score 0 · Accepted Answer · answered Jul 02 '20 at 00:17

No, addresses are not supposed to work like normal numbers. The compiler automatically adjusts address arithmetic to work in units of elements, using elements of the type pointed to.

Do not print addresses with %d. Use %p and cast the address to void *.

Do not print differences of addresses with %d. Use %td.

A correct printf would be printf("%p - %p = %td.\n", (void *) (i+1), (void *) i, (i+1) - i);. This will print “1” for the difference because the addresses i+1 and i differ by one element.

Aren't address supposed to work like normal numbers (int)?

1 Answers1