Address in C: &(number)

Question

I don't understand the output of this program:

int arr[]={1,7,4,2,5,8};
int x=(&(arr[arr[1]-arr[4]])-arr);
printf("%d" ,x);

arr[arr[1]-arr[4]] is equal to 4. What does it mean &(4)? Why does it print 2?

Answer 1

arr[1] - arr[4]

This is just as it looks. 7 - 5 = 2, so let's replace that with 2:

arr[2]

That's also just as it looks. 4. The & takes its address, which will be two ints offset from the beginning of the array, which is arr.

&(arr[2]) - arr

That subtracts arr, so you're left with the offset of arr[2] from arr, which is two ints. There you go.

Here's a reduced example.

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In case you were expecting it to be 8, well, that's just how pointer arithmetic is. Casting them both to unsigned int:

(unsigned int)&arr[2] - (unsigned int)arr

will yield 8, at least when an int is four bytes. (Demo)

Answer 2

Look at minitech's answer for showing how the print value of "2" can be deduced.

However, I think that in your case you just misunderstood how the & operator works. Yes, arr[arr[1]-arr[4]] is 4, but that does not mean that &(arr[arr[1]-arr[4]]) is doing &4; it's doing &(arr[2]); and it's taking the address of arr[2], not the value of arr[2], which is 4.