how do arithmetic operators work with addresses in c?

Question

int main(void){
   int arr[] = {1, 2, 3, 4, 5, 6};
   printf("%p\t%p\t%d\n",arr,arr+1,(arr+1)-arr);
   return 0;
 }

output :

  0x7ffe583f4ba0    0x7ffe583f4ba4  1

why is the difference 1? Shouldn't it be 4.

The difference is in units of what the pointer is pointing to. In your case in units of `sizeof *arr`. — Some programmer dude, Aug 29 '18 at 13:39
`arr + 1 - arr` = `arr - arr + 1` = `0 - 1`, why would that be any different :? — Antti Haapala -- Слава Україні, Aug 29 '18 at 13:42
Sidenote: the difference between two pointers yields a `ptrdiff_t`, not an `int`. Your conversion type specifier invokes undefined behaviour. And why do you assume `sizeof(int) == 4`? — too honest for this site, Aug 29 '18 at 13:58
@toohonestforthissite The output of the two `%p` values shared by the OP have a difference of 4 — Govind Parmar, Aug 29 '18 at 14:03

haccks · Accepted Answer · 2018-08-29T13:47:15.283

5

Think this expression (arr+1)-arr as

0x7ffe583f4ba0 + 1 - 0x7ffe583f4ba0

it will give 1.

When you subtract two pointers pointing to the same array then it gives the number of elements between those pointers.
By the same logic, if you increment a pointer to an array of int by 1 then it will point to it's next element (one unit) and not to the next sizeof(int) element.

edited Aug 29 '18 at 13:47

answered Aug 29 '18 at 13:39

haccks

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score 0 · Answer 2 · answered Aug 29 '18 at 13:41

When you perform pointer arithmetic with + and - the size of the type pointed to is taken into account:

int main() {
    int arr[2] = { 1, 2 };
    int *p = arr;
    printf("%p -> %d\n", (void *)p, *p);
    p++;
    printf("%p -> %d\n", (void *)p, *p);
    return 0;
}

Output:

0115FCCC -> 1
0115FCD0 -> 2

how do arithmetic operators work with addresses in c?

2 Answers2