将lambda作为模板参数传递给模板化的函数指针函数 [英] Passing lambda as template parameter to templated by function-pointer function

问题描述

看起来我不能将一个不捕获的lambda作为模板参数传递给一个模板化的函数指针函数。我是以错误的方式做，还是不可能？

Looks like I cannot pass a no-capture lambda as a template parameter to a templated by function-pointer function. Am I doing it the wrong way, or is it impossible?

#include <iostream>

// Function templated by function pointer
template< void(*F)(int) >
void fun( int i )
{
    F(i);
}

void f1( int i )
{
    std::cout << i << std::endl;
}

int main()
{
    void(*f2)( int ) = []( int i ) { std::cout << i << std::endl; };

    fun<f1>( 42 ); // THIS WORKS
    f2( 42 );      // THIS WORKS
    fun<f2>( 42 ); // THIS DOES NOT WORK (COMPILE-TIME ERROR) !!!

    return 0;
}

推荐答案

语言的定义，以下使它更明显：

It's mostly a problem in the language's definition, the following makes it more obvious:

using F2 = void(*)( int );

// this works:
constexpr F2 f2 = f1;

// this does not:
constexpr F2 f2 = []( int i ) { std::cout << i << std::endl; };

活动示例

这基本上意味着您的希望/期望是相当合理的，但目前语言没有定义 - lambda不会产生适合作为 constexpr 的函数指针。

This basically means that your hope/expectation is quite reasonable, but the language is currently not defined that way - a lambda does not yield a function pointer which is suitable as a constexpr.

但是，修正此问题的提案： N4487 。

There is, however, a proposal to fix this issue: N4487.

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