使用void_t的多个SFINAE类模板专业化

Question

Are multiple class template specialisations valid, when each is distinct only between patterns involving template parameters in non-deduced contexts?

A common example of std::void_t uses it to define a trait which reveals whether a type has a member typedef called "type". Here, a single specialisation is employed. This could be extended to identify say whether a type has either a member typedef called "type1", or one called "type2". The C++1z code below compiles with GCC, but not Clang. Is it legal?

template <class, class = std::void_t<>>
struct has_members : std::false_type {};

template <class T>                      
struct has_members<T, std::void_t<typename T::type1>> : std::true_type {};

template <class T>                                                        
struct has_members<T, std::void_t<typename T::type2>> : std::true_type {};

Answer 1

There is a rule that partial specializations have to be more specialized than the primary template - both of your specializations follow that rule. But there isn't a rule that states that partial specializations can never be ambiguous. It's more that - if instantiation leads to ambiguous specialization, the program is ill-formed. But that ambiguous instantiation has to happen first!

It appears that clang is suffering from CWG 1558 here and is overly eager about substituting in void for std::void_t .

This is CWG 1980 almost exactly:

In an example like

 template<typename T, typename U> using X = T; template<typename T> X<void, typename T::type> f(); template<typename T> X<void, typename T::other> f(); 

it appears that the second declaration of f is a redeclaration of the first but distinguishable by SFINAE, ie, equivalent but not functionally equivalent.

If you use the non-alias implementation of void_t :

template <class... Ts> struct make_void { using type = void; };
template <class... Ts> using void_t = typename make_void<Ts...>::type;

then clang allows the two different specializations. Sure, instantiating has_members on a type that has both type1 and type2 typedefs errors, but that's expected.

Answer 2

I don't believe it's correct, or at least, not if we instantiate has_members with a type that has both type1 and type2 nested, the result would be two specializations that are

has_members<T, void> 

which would not be valid. Until the code is instantiated I think it's ok, but clang is rejecting it early. On g++, your fails with this use-case, once instantiated:

struct X
{
    using type1 = int;
    using type2 = double;
};

int main() {
    has_members<X>::value;
}

The error message is doesn't seem to describe the actual problem, but it at least is emitted:

<source>:20:21: error: incomplete type 'has_members<X>' used in nested name specifier
     has_members<X>::value;
                     ^~~~~

If you instantiate it with a type that has only type1 or type2 but not both, then g++ compiles it cleanly. So it's objecting to the fact that the members are both present, causing conflicting instantiations of the template.

To get the disjunction, I think you'd want code like this:

template <class, class = std::void_t<>>
struct has_members : std::bool_constant<false> {};

template <class T>
struct has_members<T, std::enable_if_t<
        std::disjunction<has_member_type1<T>, has_member_type2<T>>::value>> : 
    std::bool_constant<true> {};

This assumes you have traits to determine has_member_type1 and has_member_type2 already written.