How does subtracting an address reference work?

Question

I think the value of z should be 40 because a[5] has 20 elements, and the space between it and a[3] also has 20 elements. However, the actual value of z is 2.

Can anyone explain this concept?

#include <stdio.h>

int main()
{
    int a[10][20];
    int z = &a[5] - &a[3];
    printf("%d\n", &a[3]); // the address information is 6421740
    printf("%d\n", &a[5]); // the address information is 6421900
    printf("%d\n", z);     // z value is 2. why?
}

Answer 1

Pointer arithmetic is done in the units of the pointed-to type.

In this case a[5] and a[3] are both elements of an array of type int [20], and &a[5] and &a[3] both have type int(*)[20]. They are 2 array element apart, so the difference between them is 2.

It doesn't matter that the underlying type is also an array.