从重载函数中提取返回类型 [英] Extracting the return type from an overloaded function

问题描述

我想提取函数的返回类型。问题是，有其他功能具有相同的名称，但不同的签名，我不能让C ++选择适当的。我知道std :: result_of，但从几个尝试，我得出结论，它也遭受同样的问题，以及。我已经听说过一个涉及decltype的解决方案，但我不知道任何细节。

I want to extract the return type of a function. Problem is, there are other functions with the same name but different signature, and I can not get C++ to select the appropriate one. I know about std::result_of, but from a few tries I have concluded it suffers from the same problem as well. I have heard about a solution involving decltype as well, but I do not know any specifics.

目前我使用模板元编程从一个函数中提取返回类型指针类型，对于有限数量的参数（任何非限制性解决方案？）都可以正常工作，假定函数指针类型的提取对于无歧义函数起作用。

At the moment I am using template metaprogramming to extract the return type from a function pointer type, which works fine for a limited number of parameters (any non-limited solution?), given that extraction of function pointer type works for unambiguous functions.

#include <iostream>

using namespace std;

//  ----

#define resultof(x)     typename ResultOf<typeof(x)>::Type  //  might need a & before x

template <class T>
class ResultOf
{
    public:
        typedef void Type;      //  might need to be T instead of void; see below
};

template <class R>
class ResultOf<R (*) ()>
{
    public:
        typedef R Type;
};

template <class R, class P>
class ResultOf<R (*) (P)>
{
    public:
        typedef R Type;
};

//  ----

class NoDefaultConstructor
{
    public:
        NoDefaultConstructor (int) {}
};


int f ();
int f ()
{
    cout << "f" << endl;
    return 1;
}

double f (int x);
double f (int x)
{
    cout << "f(int)" << endl;
    return x + 2.0;
}

bool f (NoDefaultConstructor);
bool f (NoDefaultConstructor)
{
    cout << "f(const NoDefaultConstructor)" << endl;
    return false;
}

int g ();
int g ()
{
    cout << "g" << endl;
    return 4;
}

int main (int argc, char* argv[])
{
    if(argc||argv){}

//  this works since there is no ambiguity. does not work without &
//  resultof(&g) x0 = 1;
//  cout << x0 << endl;

//  does not work since type of f is unknown due to ambiguity. same thing without &
//  resultof(&f) x1 = 1;
//  cout << x1 << endl;

//  does not work since typeof(f()) is int, not a member function pointer; we COULD use T instead of void in the unspecialized class template to make it work. same thing with &
//  resultof(f()) x2 = 1;
//  cout << x2 << endl;

//  does not work per above, and compiler thinks differently from a human about f(int); no idea how to make it correct
//  resultof(f(int)) x3 = 1;
//  cout << x3 << endl;

//  does not work per case 2
//  resultof(f(int())) x4 = 1;
//  cout << x4 << endl;

//  does not work per case 2, and due to the lack of a default constructor
//  resultof(f(NoDefaultConstructor())) x5 = 1;
//  cout << x5 << endl;

//  this works but it does not solve the problem, we need to extract return type from a particular function, not a function type
//  resultof(int(*)(int)) x6 = 1;
//  cout << x6 << endl;

}

任何想法我缺少什么语法特性，它优选地具有以简单方式工作的解决方案，例如， resultof（f（int））？

Any idea what syntax feature am I missing and how to fix it, preferably with a solution that works in a simple way, e.g. resultof(f(int))?

推荐答案

可以使用 decltype 和 declval ：

例如： decltype（f（std :: declval< T>（）））。

这篇关于从重载函数中提取返回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！