C++11 在成员函数上应用 result_of，失败，为什么？

Question

我有以下代码作为实验：

int f1() { return 0; }

struct Bar {
    Bar() = delete;
    int f() { return 0; }
    int operator()() { return 1; }
};

int main()
{
    decltype(f1()) x = 3;//f1() is expression
    result_of<decltype(&f1)()>::type x1 = 3;//type+param
    result_of<Bar()>::type x3 = 3;//type+param
    decltype(declval<Bar>().f()) y = 4;//expression
    decltype((((Bar*)nullptr)->*(&Bar::f))()) z = 5;//expression

    result_of<decltype(std::mem_fn(&Bar::f))()>::type y2 = 3;//error!!!!!!
}

除了最后一个result_of，一切正常：我试图使用 result_of.

获取 return 类型的 Bar::f

为什么失败，如何纠正？

Answer 1

未指定return类型mem_fn:

template <class R, class T>
unspecified mem_fn(R T::* pm) noexcept;

是根据 INVOKE [func.memfn]/p1:

定义的

¹ Returns: A simple call wrapper ([func.def]) fn such that the expression fn(t, a2, ..., aN) is equivalent to INVOKE(pm, t, a2, ..., aN) ([func.require]).

其中 INVOKE 的定义包括以下两个项目符号 [func.require]/p1:

Define <I>INVOKE</I>(f, t1, t2, ..., tN) as follows:

— (t1.*f)(t2, ..., tN) when f is a pointer to a member function of a class T and is_base_of<T, decay_t<decltype(t1)>>::value is true;

— ((*t1).*f)(t2, ..., tN) when f is a pointer to a member function of a class T and t1 does not satisfy the previous two items;

即what mem_fn return的第一个参数必须是隐式对象参数（t1）的类型，可以是引用也可以是指针，例如：

std::result_of<decltype(std::mem_fn(&Bar::f))(Bar&)>::type y2;
//                                            ~~~^

std::result_of<decltype(std::mem_fn(&Bar::f))(Bar*)>::type y2;
//                                            ~~~^

您也可以完全删除 std::mem_fn：

std::result_of<decltype(&Bar::f)(Bar*)>::type y2;

C++11 在成员函数上应用 result_of，失败，为什么？

C++11 applies result_of on member function, failed, why?

c++

types

result-of

decltype

c++11