Should conversion operators be considered for function template argument deduction?

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Consider the following code:

template<typename>
struct S 
{
    operator S<int&>();  
};

template<typename T>
void f(S<T&>);

int main() 
{
    f(S<int&&>{});  // gcc ok
                    // clang error
}

gcc uses the conversion operator on the temporary argument, returning S<int&> which is matched by S<T&>, and accepts the call.

clang doesn't consider the conversion operator, fails to match T& against int&&, and rejects the call.

So what does the language say should happen here?

template-argument-deduction language-lawyer templates c++
1 Answers

GCC is certainly wrong here: T& and T&& are different rows in [temp.deduct.type]/8 and are thus incompatible. Why it does so is not clear. It would make more sense to err in the other direction: if the parameter were declared as S<T&&> and the argument were of type S<int&>, there would at least be a T (i.e., int&) such that (due to reference collapsing) the parameter and argument types were the same. (There would also be the easy mistake to make to say that a universal reference was involved.)

user_85862270
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