C++中获取返回值的类型

假设我们有一个函数f它返回一个未知类型的值（我们称之为T) 并采用类型的值T作为一个论点（可能还有一些其他论点）。我如何获得返回类型f在 C++14 中？

如果我们知道参数类型，就有一种方法可以做到这一点（通过std::result_of）。如果我们知道除T?

例子：

template <class F> // F is functor with   T operator()(T a, T b)
class A {
    // Here I want to do
    // T some_function(T some_arg) { ... }
}

template <typename T>
struct return_type;

template <typename R, typename... Args>
struct return_type<R(Args...)> { using type = R; };

template <typename R, typename... Args>
struct return_type<R(*)(Args...)> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type<R(C::*)(Args...)> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type<R(C::*)(Args...) &> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type<R(C::*)(Args...) &&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type<R(C::*)(Args...) const> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type<R(C::*)(Args...) const&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type<R(C::*)(Args...) const&&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type<R(C::*)(Args...) volatile> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type<R(C::*)(Args...) volatile&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type<R(C::*)(Args...) volatile&&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type<R(C::*)(Args...) const volatile> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type<R(C::*)(Args...) const volatile&> { using type = R; };

template <typename R, typename C, typename... Args>
struct return_type<R(C::*)(Args...) const volatile&&> { using type = R; };

template <typename T>
using return_type_t = typename return_type<T>::type;

Test:

#include <type_traits>

struct Functor
{
    int operator()(int i, int j) { return i + j; }
};

template <class F>
struct A
{
    using T = return_type_t<decltype(&F::operator())>;

    T some_function(T some_arg) { return some_arg; }
};

int main()
{
    A<Functor> a;
    static_assert(std::is_same<decltype(a.some_function(1)), int>::value, "!");
}

DEMO