c++ - 为什么不调用在构造函数中作为参数传递的自由函数？

Question

为什么 mystruct( plain_old_function ); 没有构造函数调用默认构造函数，而与 lambda 一样调用专用构造函数 (mystruct ( const std::function< std::string() > &func ))？

这可以工作吗？

#include <iostream>
#include <functional>
#include <string>

struct mystruct
{
    mystruct() { std::cout << "Default construct :S" << std::endl; }
    mystruct ( const std::function< std::string() > &func )  {
        std::cout << func() << std::endl;
    }
};

void callme ( const std::function< std::string() > &func )
{
    std::cout << func() << std::endl;
}

std::string free_function(  ) {  return "* Free function"; }


int main()
{

    std::cout << "Constructing with lambda:" << std::endl;
    mystruct( [](){ return "* Lambda function"; } );

    std::cout << "Calling free  function through another function:" << std::endl;
    callme( free_function );

    std::cout << "Constructing with free function:" << std::endl;
    mystruct( free_function );

    return 0;
}

Demo

输出:

Constructing with lambda:
* Lambda function
Calling free  function through another function:
* Free function
Constructing with free function:
Default construct :S

Answer 1

烦人的解析，

mystruct( free_function );

被解析为

mystruct free_function; // declare a mystruct instance named free_function
                        // (hiding the function)

您可以使用 {}:

mystruct{free_function};