Why does conceptual storage allocation differ from the actual?

Question

I have a puzzling question (at least for me)

Say I declare an integer array:

int arr[3];

Conceptually, what happens in the memory is that, at compile time, 12 bytes are allocated to store 3 consecutive integers, right? (Here's an illustration)

Based on the illustration, the sample addresses of

arr[0] is 1000,

arr[1] is 1004, and

arr[2] is 1008.

My question is:

If I output the difference between the addresses of arr[0] and arr[1]:

std::cout << &arr[1] - &arr[0] << std::endl;

instead of getting 4,

I surprisingly get 1.

Can anybody explain why it resulted to that output?

PS: On my computer, an int is 4 bytes.

It's pointer arithmetic, you'll receive the number of elements not bytes. — πάντα ῥεῖ, May 14 '16 at 14:58
Sorry, but I didn't quite get it. Dealing with addresses (like adding, subtracting, multiplying, diviing) is dealing with bytes, no? — Christian, May 14 '16 at 15:06
_@Christian_ No. Note these are typed addresses, not like `void*`. — πάντα ῥεῖ, May 14 '16 at 15:07
@πάντα ῥεῖ I get it now :D thanks for your effort! — Christian, May 14 '16 at 15:16

score 0 · Answer 1 · answered May 14 '16 at 14:58

Pointer arithmetic automatically divides the value by the size of the base type so this is not surprising at all since one would expect to get 4 / 4 which is 1. Cast to unsignd char * to see the difference.

#include <iostream>

int
main(void)
{
    int arr[2];

    std::cout << &arr[1] - &arr[0] << std::endl;
    std::cout << reinterpret_cast<unsigned char *>(&arr[1]) -
        reinterpret_cast<unsigned char *>(&arr[0]) << std::endl;

    return 0;
}

Why does conceptual storage allocation differ from the actual?

I have a puzzling question (at least for me)

1 Answers1