Why doesn't std::extent work on references to arrays like operator sizeof?

Question

If I want to get the size (number of elements) of an array I can use sizeof(a) / sizeof(a[0]) but the new standard makes it possible to use type traits to do that:

int main(){

    int a[]{5, 7, 2, 3, 6, 7};

    std::cout << std::extent<decltype(a)>::value << '\n'; // 6

    auto& refA = a;
    std::cout << std::extent<decltype(refA)>::value << '\n'; // 0
    std::cout << sizeof(refA) / sizeof(refA[0]) << '\n'; // 6
    std::cout << std::extent<std::remove_reference<decltype(refA)>::type>::value << '\n'; // 6
 
    std::cout << "\ndone!\n";
}

Everything is OK but why doesn't std::extent work on references to arrays as the sizeof() operator does? I've had to remove the reference in the last example to get the actual array type.

Answer 1

This doesn't answer the question as written directly, but rather a potential xy problem. You shouldn't nor do you need to use std::extent to do this. It's not useful in this case. Use std::size instead:

std::size(a)

Now for the answer:

why doesn't std::extent work on references to arrays

It works; just not in the way you may have been expecting. In particular, the extent of any reference is 0 rather than the extent of the referred type.