Why can't it deduce the class type?

[LB]

http://ideone.com/DbOImJ

#include <iostream>
#include <type_traits>

struct Test
{
	void f();
	template<typename T>
	using verify = typename std::enable_if<std::is_same<T, T>::value, T>::type;
	template<typename T, typename R, typename... Args>
	using MFP = R (verify<T>::*)(Args...);
	template<typename T, typename R, typename... Args>
	static MFP<T, R, Args...> g(MFP<T, R, Args...> mf){return mf;}
};

int main()
{
	auto test = Test::g(&Test::f);
}

prog.cpp: In function ‘int main()’:
prog.cpp:17:30: error: no matching function for call to ‘Test::g(void (Test::*)())’
  auto test = Test::g(&Test::f);
                              ^
prog.cpp:17:30: note: candidate is:
prog.cpp:12:28: note: template<class T, class R, class ... Args> static R (std::enable_if<std::is_same<_Tp, _Tp>::value, T>::type::* Test::g(Test::MFP<T, R, Args ...>))(Args ...)
  static MFP<T, R, Args...> g(MFP<T, R, Args...> mf){return mf;}
                            ^
prog.cpp:12:28: note:   template argument deduction/substitution failed:
prog.cpp:17:30: note:   couldn't deduce template parameter ‘T’
  auto test = Test::g(&Test::f);
                              ^

Obviously in my real code I am not using std::is_same<T, T>, I'm actually using std::is_base_of, but the issue is that if I use verify at all the compiler fails to deduce the class type.

[o1vk4iN6]

It is trying to match "Test" with "verify<T>" and it can't deduce the type of T from that (the type can't be deduced until the value of T is know so it may not equal "void" [such in the case where it fails]). You could try moving the verification outside of the g function.

http://ideone.com/VGTszV

[LB]

Thanks, I came up with a solution similar to this.