How *(&arr + 1) - arr is working to give the array size [duplicate]

int arr[] = { 3, 5, 9, 2, 8, 10, 11 };      
int arrSize = *(&arr + 1) - arr;
std::cout << arrSize;

I am not able to get how this is working. So anyone can help me with this.

edited May 12 at 19:13

largest_prime_is_463035818

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asked May 12 at 19:02

Rahul Goswami

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25

This is not a usual idiom. Other readers of the code will be confused as well. Sure, it's a clever trick. Avoid cleverness like the plague. – dratenik May 12 at 19:14
16

Just use std::size – Ayxan Haqverdili May 12 at 19:15
2

@ThomasMatthews I was curious, but I dont think we can learn much from this: godbolt.org/z/YdWh8h8YY – largest_prime_is_463035818 May 12 at 19:30
3

This is the same question but for C: stackoverflow.com/q/56154380/1896169 The C and C++ standards are sufficiently different, however, that it makes sense to have separate questions – Justin May 13 at 4:12
8

FWIW, another alternative is sizeof(arr) / sizeof(arr[0]), IIRC. – Solomon Ucko May 13 at 17:42

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