如何拥有指向通用lambda的函数指针? [英] How to have a function pointer to a generic lambda?

问题描述

正如标题所述，如何在代码中表达以下意图?作为函数指针的要求采用任何类型的参数.由于 std :: string .Variadic无法正常工作.

As the title says, how can I have the following intent expressed in code ? The requirements being the function pointer takes any type of arguments. Variadic doesn't work because of std::string.

https://godbolt.org/z/1E1szT

请注意，由于 fp 是类成员变量，因此我不能直接使用 auto fp =< the lambda> .

Note that I cannot directly use auto fp = <the lambda> because fp is a class member variable.

#include <iostream>

template<typename T1, typename T2>
void(*fp)(T1 t1, T2 t2) = [](auto a, auto b){
                              std::cout << a << "--"
                              << b << std::endl;
                          };

int main(){
    fp(1, 2);
    fp('a', 'b');
}

推荐答案

变量模板非常好.
甚至您初始化它们的方式都很好.

A variable template is perfectly fine.
Even the way you initialize them is fine.

请注意，它是变量的模板，而不是存储模板(不存在)的变量.

Just be aware it is a template for variables, not a variable storing a template (which cannot exist).

因此，当最终使用它时，事情就崩溃了:

Thus when finally using it, things fall apart:

编译器应如何知道您指的是无数个潜在实例?
很简单，您必须实际上是告诉它 :

How should the compiler know which of the myriad potential instances you mean?
Simple, you have to actually tell it:

int main(){
    fp<int, int>(1, 2);
    fp<char, char>('a', 'b');
}

当然，手动对lambda进行减半运算并获取其实例会更实用:

Of course, manually desugaring the lambda and getting an instance of it would be more practical:

struct {
    template <class T1, class T2>
    void operator()(T1 a, T2 b) const {
        std::cout << a << "--" << b << std::endl;
    };
} fp;

遗憾的是，我们不能简单地给lambda一个名称并让编译器弄清楚.
为合适的语法集思广益:

A shame we cannot simply give a lambda a name and let the compiler figure it out.
Brainstorming candidates for appropriate syntax:

struct A = []...;
struct B : []... {
};
using C = []...;

这篇关于如何拥有指向通用lambda的函数指针?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！