C ++ 11-模板std :: enable_if和std :: result_of

Question

I have created this to specialize template for void / non-void methods

template <typename ClassType, 
        typename MethodType, MethodType MethodName,         
        typename std::enable_if <std::is_same<void, std::result_of<decltype(MethodName)(ClassType)>::type>::value> ::type* = nullptr
>
static int function()
{
   //void
   //....
}



template <typename ClassType, 
        typename MethodType, MethodType MethodName,         
        typename std::enable_if <!std::is_same<void, std::result_of<decltype(MethodName)(ClassType)>::type>::value> ::type* = nullptr
>
static int function()
{
   //non-void
   //....
}

//And I want to call it like this
function<Foo, void (Foo::*)(void), &Foo::Print>();
function<Foo, int (Foo::*)(void), &Foo::Print2>();

(Based on this answer: C++ template parameter as function call name )

Hovewer, this gives me bunch of errors (MSVC 2015). If I run this inside of

template <typename ClassType, 
        typename MethodType, MethodType MethodName      
>
static int print()
{
    std::cout << "C: " << std::is_same<void, std::result_of<decltype(MethodName)(ClassType)>::type>::value << std::endl;
}

I got true result.

Is it possible to "create" function with specialisation for void / non-void results of MethodName ?

Answer 1

This compiles fine under GCC

#include <iostream>
#include <type_traits>
using namespace std;

template <typename ClassType, typename MethodType, MethodType MethodName>
static auto function()
  -> std::enable_if_t<std::is_void<typename std::result_of<MethodType(ClassType)>::type>::value, int>
{
   //void
   //....
}

template <typename ClassType,  typename MethodType, MethodType MethodName>
static auto function()
  -> std::enable_if_t<!std::is_void<typename std::result_of<MethodType(ClassType)>::type>::value, int>
{
   //non-void
   //....
}

I am not sure if this is what you are looking after but I moved enable_if to return type using the arrow syntax, this looks just more clean to me. Also why bother using decltype on MethodName since you already have the type as MethodType. Also result_of needs typename before accessing the type.

This as above does compile although without possible usage I am not sure if this is what you are after.

note: std::enable_if_t is avaialble in C++14 if you can't use that change to typename std::enable_if<...>::type instead.

Answer 2

You have just miss typename .

As alternative I suggest:

template <typename F, F f> struct function_helper;

template <typename C, typename ... Ts, void (C::*m)(Ts...)>
struct function_helper<void (C::*)(Ts...), m>
{
    int operator()() const { /* void implementation*/ }
};

template <typename C, typename Ret, typename ... Ts, void (C::*m)(Ts...)>
struct function_helper<Ret (C::*)(Ts...), m>
{
    int operator()() const { /* non-void implementation*/ }
};

template <typename F, F f>
static int function()
{
     return function_helper<F, f>{}();
}

With usage

function<void (Foo::*)(), &Foo::Print>();
function<int (Foo::*)(), &Foo::Print2>();

or

function<decltype(&Foo::Print), &Foo::Print>();
function<decltype(&Foo::Print2), &Foo::Print2>();

With C++17, we may get rid of first template argument with template <auto f> struct function_helper;