如何在C ++ 11中为result_of编写帮助类型 [英] How to write a helper type for result_of in C++11

问题描述

template<typename F,typename...X>
using Result_of = typename result_of<F(X...)>::type;

int ff(int){return 2;}
typedef bool(*PF)(int);
auto fx = [](char ch){return tolower(ch);};
Result_of<decltype(&ff)()> r1 = 7;
Result_of<PF(int)> r2 = 2;
Result_of<decltype(fx)(char)> r5 = "a";

当我用gcc编译时，得到以下错误：

When I compile with gcc, I get the following error:

main.cpp: In substitution of 'template<class F, class ... X> using Result_of = typename std::result_of<_Functor(_ArgTypes ...)>::type [with F = int (*())(int); X = {}]':
main.cpp:17:30:   required from here
main.cpp:6:57: error: function returning a function
 using Result_of = typename std::result_of<F(X...)>::type;
                                                         ^

为什么我收到此错误，如何解决？ >

Why am I getting this error and how do I fix it?

推荐答案

这行有几个问题：

Result_of<decltype(&ff)()> r1 = 7;

首先， Result_of 采用函数类型参数列表，以逗号分隔。你不是这样提供的。因此，它将 F 解释为（int）（*）（int）（）一个函数 int 返回一个返回 int 的函数。但是在C ++中有一个函数返回一个函数是非法的，因此错误。

First, Result_of takes a function type and a list of arguments, comma separated. You're not providing them that way. So it's interpreting F to be (int)(*)(int)(), which would be a pointer to a function taking an int returning a function returning an int. But it's illegal in C++ to have a function return a function, hence the error.

其次， ff 接受 int 。最后， ff 衰减到指针函数，您需要实际的函数。正确的表达式是：

Second, ff takes an int, not nothing. Lastly, ff decays to a pointer-to-function and you need the actual function. The correct expression would be:

Result_of<decltype(*ff), int> r1 = 7;

同样，下一个需要是：

Result_of<PF, int> r2 = 2;
Result_of<decltype(fx), char> r5 = 'a'; // fx returns an int, so you 
                                        // can't assign a const char* to it

这篇关于如何在C ++ 11中为result_of编写帮助类型的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！