使用std :: function的模板替换失败 [英] Template substitution failure with std::function

问题描述

我试图将回调函数作为函数参数传递。但是在以下代码中出现模板替换失败错误。不确定为什么模板替换失败。

I am trying to pass a callback function as function parameter. But getting template substitution failure errors in following code. Not sure why template substitution is failing.

#include<iostream>
#include <map>
#include <tuple>
#include <functional>

template<typename A,typename B>
void myfun(std::map<A,B> & mm, std::function<std::tuple<A,B>(void)> fn)
{
    A key;
    B val;
    std::tie(key,val) = fn();
    mm[key] = val;
}

std::tuple<std::string,int> fun()
{
    return std::make_tuple(std::string("hi"),1);
}

int main()
{
    std::map<std::string,int> gg;
#if 0
        //fixed version
        std::function<std::tuple<std::string,int>(void)> yy = fun;//fixed
        myfun(gg,yy);//fixed
#else
        // error causing code
        myfun(gg,fun);
#endif
}

错误如下

main.cpp:8:6: note:   template argument deduction/substitution failed:

main.cpp:25:17: note:   mismatched types 'std::function<std::tuple<_T1, _T2>()>' and 'std::tuple<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >, int> (*)()'

     myfun(gg,fun);

推荐答案

编译器不能同时转换为 std :: function 并推导出模板参数。它不理解任意函数指针和 std :: function 之间的映射。

The compiler can't both cast to a std::function and deduce the template arguments. It doesn't understand the mapping between an arbitrary function pointer and a std::function.

有一个

您可以在呼叫站点上显式创建 std :: function ：

You could explicitly create a std::function at the call site:

 myfun(gg,std::function<std::tuple<std::string,int>(void)>{fun});`

您可以编写 make_function 函数为您推断类型。您可以在网上找到有关此主题的讨论和实现，例如在此处，此处和

You could write a make_function function to deduce the types for you. You can find discussions and implementations of this online, such as here, here and here.

myfun(gg,make_function(fun));

您可能会忘记 std :: function 并推导出整个函数类型。这是我要采用的方法：

You could just forget about std::function and deduce the entire function type. This is the approach I would take:

template<typename A,typename B, typename Fun>
void myfun(std::map<A,B> & mm, Fun fn)
{
    A key;
    B val;
    std::tie(key,val) = fn();
    mm[key] = val;
}

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