Find size of array without using sizeof

&arr gives you a pointer to the array. (&arr)[1] is equivalent to *(&arr + 1). &arr + 1 gives you a pointer to the array of 100 ints that follows arr. Dereferencing it with * gives you that array that follows. Since this array is used in an additive expression (-), it decays to the pointer to its first element. The same happens to arr in the expression. So you subtract to pointers, one pointing to the non-existent element right after arr and the other pointing to the first element of arr. This gives you 100.

But it's not working. %d is used for int. Pointer difference returns you ptrdiff_t and not int. You need to use %td for ptrdiff_t. If you lie to printf() about the types of the parameters you're passing to it, you get well-deserved undefined behavior.

EDIT: (&arr)[1] may cause undefined behavior. It's not entirely clear. See the comments below, if interested.

Generally (as per visual studio), for an array &arr is same as arr ,which return the starting base address of our function.

(&arr)[0] is nothing but &arr or arr

ex: it will return some address : 1638116

Now, (&arr)[1] means we are started accessing the array out of bounce means next array or next segment of the size of present array(100 ahead).

ex: it will return some address : 1638216

Now, subtracting (&arr)[1] - (&arr)[0]=100

&arr is a pointer to an array of 100 ints.

The [1] means "add the size of the thing that is pointed to", which is an array of 100 ints.

So the difference between (&arr)[1] and arr is 100 ints.

(Note that this trick will only work in places where sizeof would have worked anyway.)