Pointers & Arrays

Array name = address of element 0

In most expressions, an array's name decays into a pointer to its first element. That's why arr and &arr[0] are interchangeable, and arr[i] is really *(arr + i).

Same result, two notations

Indexing and pointer-offset dereferencing give identical results.

int arr[4] = {1, 2, 3, 4};
printf("%d\n", arr[2]);
printf("%d\n", *(arr + 2));  // same thing

Output:

3
3

But arrays aren't pointers

sizeof(arr) gives the whole array's byte size, while sizeof on a pointer gives just the pointer's size (usually 8 bytes). Decay only happens when the array is used, not when it's sizeof'd.