pointer arithmetic (char*) &a[1] - (char *)&a[0] == 4

Question

If a is an int array, (char*) &a[1] - (char *)&a[0] is equal to 4, while &a[1] - &a[0] is equal to 1. why is that?

Answer 1

Pointer math operates on the size of the data structure its pointing to. This is because if I do this:

int array[10] ;
int * p = array ;

p ++ ;

I want p pointing at the second int, not some memory halfway in between two elements.

So &a[1] is four bytes apart from &a[0] but asking it &a[1] - &a[0] asks how many ints apart it is. When you cast it to char you ask for the math in terms of the size of char.

Answer 2

When you do

&a[1] - &a[0]

since a is an int array, an implicit (int *) pointer is assumed, that is

(int *)&a[1] - (int *)&a[0]

Hence since both are of type pointer to int , their difference gives 1.

But when you do-

(char*) &a[1] - (char *)&a[2]

assuming int is 4 bytes and char is 1 byte on your compiler, the difference will be four since each element of a is int and has four bytes.