Why cannot I find the element size via this pointer hack? [duplicate]

Question

Consider the following in C

int arr[]= {1,2,3,4,5};

My system uses 4 bytes to store int data. Now &arr[0] => 0022FEB0 and &arr[1]=>0022FEB4 at a moment of run

Now

int diff=&arr[1]-&arr[0];

Value stored in diff is 1 and not 4.

Why?

Answer 1

That's the way pointers work. You are not calculating the byte difference. You're calculating the difference in number of elements.

To get the element size, use sizeof(*arr)

To get the byte difference, use (&arr[1]-&arr[0]) * sizeof(*arr)

Answer 2

I think that what you want it this:

int diff = (&arr[1]-&arr[0]) * sizeof(int);

When you tells to compiler the data type you are working with, it will consider it in the arithmetic.

Answer 3

You are not calculating the byte difference. To calculate byte difference there is already an answer on StackOverflow. here is a link to answer Getting the difference between two memory addresses

Answer 4

The compiler knows that types of substracted ones whose are pointer to int.

#include <stdio.h>
#include <stddef.h>

int main(void) {

    int arr[]= {1,2,3,4,5};

    /*
    int diff=&arr[1]-&arr[0];

    In your system, there's a 4 byte difference in memory, so you'd get diff = 1
    here because the compiler knows they're ints so the result is diff/sizeof(int)
    */

    uintptr_t firstNumAddress  = (uintptr_t)&arr[0];
    uintptr_t secondNumAddress = (uintptr_t)&arr[1];

    ptrdiff_t ptrDiff = secondNumAddress - firstNumAddress;
    printf("%lu - %lu = %ti\n", secondNumAddress, firstNumAddress, ptrDiff);

    return 0;
}