CWG2978 [temp.deduct.call] Deduction and qualification conversions with references

[t3nsor]

Full name of submitter: Brian Bi

Issue description: There is implementation divergence in the treatment of the following deduction:

#include <cstddef>

template <std::size_t N>
void f(const int* const (&)[N]);

int main() {
    int* a[3];
    f(a);
}

It is not clear whether bullet 4.1 of [temp.deduct.call] allows qualification conversions.

Suggested resolution: Edit [temp.deduct.call]/4.1 as follows:

If the original P is a reference type, the deduced A (i.e., the type referred to by the reference) can be more cv-qualified a type other than the transformed A that is reference-compatible ([dcl.init.ref]) with the transformed A.

(We might not need to say "transformed", actually. As far as I can tell, when the original P is a reference type, we transform A only in the case where P is a forwarding reference, and in that case no difference in cv-qualification is possible.)

[eisenwave]

Note that the following code indirectly reproduces this implementation divergence, and that is how the issue was originally discovered:

#include <span>

int* pointers[3];
std::span<const int* const> s = pointers;

GCC accepts this, and Clang (with libstdc++) rejects it. See https://stackoverflow.com/q/79264902/5740428

Clang fails to call this overload:

template <size_t _ArrayExtent>
span(const int * const (&)[_ArrayExtent])

Any resolution should make this construction of std::span valid, in my opinion.

[jensmaurer]

"reference-compatible" allows derived-to-base conversions, which we don't want here (I think).

Maybe "... can be a type similar to the transformed A that is reference-compatible with the transformed A".

[jensmaurer]

CWG2978