c++ : syntax for is_member_function_pointer in a template declaration

Question

I have a template with a declaration similar to this:

template <typename Arg0, typename... Args>
class blah {};

I have two versions of the template, and I want to use one when Arg0 is a member function pointer, otherwise use the other one. I'm trying to use std::enable_if and std::is_member_function_pointer but I cannot find the correct syntax. This is what I have for the true case:

template<typename = typename std::enable_if< std::is_member_function_pointer<Arg0> >::type, typename... Args>
class blah() {}

But this obviously isn't syntactically correct.

Answer 1

When using the Boolean predicates with classes there are generally two approaches I use to make the choice:

1. If I just need to choose between two types, I use sonething like

   typename std::conditional<
       std::is_member_function_pointer<F>::value,
           type_when_true, type_when_false>::type
   

2. If things need to change more than that I derive from a base which is specialized on a Boolean covering the two implementation choices:

   template <bool, typename...>
   struct helper;
   
   template <typename... A>
   struct helper<true, A...> {
       // implementation 1
   };
   template <typename... A>
   struct helper<false, A...> {
       // the other 1
   };
   template <typename F, typename... A>
   struct actual
       : helper<std::is_member_function_pointer<F>::value, F, A...>
   {
       // typedefs, using ctors, common code, etc.
   };
   

Answer 2

Maybe "ordinary" partial specialization is sufficient?

template<class Arg0>
struct blah { bool value = false; };

template<class Ret, class C, class... Args>
struct blah < Ret (C::*)(Args...) >
{ bool value = true; };

struct test
{
    int foo(double&);
};

#include <iostream>
#include <iomanip>
int main()
{
    std::cout << std::boolalpha;
    std::cout << blah<decltype(&test::foo)>().value << std::endl;
    std::cout << blah<int>().value << std::endl;
}