why is subtracting an array from lower_bound return a dereferenced value?

Question

  int arr[] = {0, 1, 15, 22, 100, 101, 232, 2323};
  int n = sizeof(arr) / sizeof(arr[0]);
  int x = 232;

  auto a = lower_bound(arr, arr + n, x) - arr;
  if (a < n && arr[a] == x) {
    // x found at index a
    cout << x << " is found " << "at index: " << a << endl;
  }

lower_bound returns an address of the element but if we subtract - arr from it, it dereferences the pointer. How come that happens?

I'm a little confused. subtracting 2 pointers gives an int. Where do you think anything is getting dereferenced? Perhaps this https://stackoverflow.com/questions/30123334 will help. — cigien, Dec 03 '20 at 15:42

score 5 · Accepted Answer · answered Dec 03 '20 at 15:44

5

arr is an array, and when an array is used in an expression, it can decay into a pointer. Since lower_bound returns a pointer (that's the iterator type for a raw array), and arr decays into a pointer, you are subtracting the address of the found object from the address of the start of the array which gives you the number of elements between them. There is no dereferencing happening at all.

answered Dec 03 '20 at 15:44

NathanOliver

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It is actually opposite address of the first from address of the found. – Slava Dec 03 '20 at 15:47

why is subtracting an array from lower_bound return a dereferenced value?

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