arrays - Why does i[arr] work as well as arr[i] in C with larger data types?

Question

It's fairly common knowledge that if you access an element of an array as arr[i] in C that you can also access the element as i[arr], because these just boil down to *(arr + i) and addition is commutative. My question is why this works for data types larger than char, because sizeof(char) is 1, and to me this should advance the pointer by just one char.

Perhaps this example makes it clearer:

#include <string.h>
#include <stdio.h>

struct large { char data[1024]; };

int main( int argc, char * argv[] )
{
    struct large arr[4];
    memset( arr, 0, sizeof( arr ) );

    printf( "%lu
", sizeof( arr ) ); // prints 4096

    strcpy( arr[0].data, "should not be accessed (and is not)" );
    strcpy( arr[1].data, "Hello world!" );

    printf( "%s, %s, %s
", arr[1].data, 1[arr].data, (*(arr+1)).data );
    // prints Hello world!, Hello world!, Hello world!
    // I expected `hold not be accessed (and is not)' for #3 at least

    return 0;

}

So why does adding one to the array pointer advance it by sizeof( struct large )?

Answer 1

In C, pointer arithmetic is defined so that writing

ptr + k

does not advance the pointer by k bytes, but by k objects. Thus if you have a pointer to an integer array and writing

*(myIntArrayPointer + 3)

You are dereferencing a pointer to the element at index 3 in the array, not the integer that starts three bytes past the start of the object.

Similarly, if you subtract two pointers, you get the logical number of elements in-between them, not the total number of bytes. Thus writing

(myIntArrayPointer + 3) - myIntArrayPointer

yields the value 3, even though there are 3 * sizeof(int) bytes in-between them.

Hope this helps!