解释std :: ostream :: operator <<带有void函数时的行为

Question

一个好奇的问题。

以下简短程序在屏幕上打印出1。我想知道如何解释这种行为。

#include <iostream>
#include <string>

void foo() {}


int main()
{
    std::cout << foo;
}

为便于参考，我在下面列出了cplusplus上的std::ostream::operator<<签名。

但是我不太了解哪个重载版本用于功能（操作符）匹配，并且为什么打印1 。请让我知道您的想法。

arithmetic types (1)    

ostream& operator<< (bool val);
ostream& operator<< (short val);
ostream& operator<< (unsigned short val);
ostream& operator<< (int val);
ostream& operator<< (unsigned int val);
ostream& operator<< (long val);
ostream& operator<< (unsigned long val);
ostream& operator<< (long long val);
ostream& operator<< (unsigned long long val);
ostream& operator<< (float val);
ostream& operator<< (double val);
ostream& operator<< (long double val);
ostream& operator<< (void* val);

stream buffers (2)  

ostream& operator<< (streambuf* sb );

manipulators (3)    

ostream& operator<< (ostream& (*pf)(ostream&));
ostream& operator<< (ios& (*pf)(ios&));
ostream& operator<< (ios_base& (*pf)(ios_base&));

顺便说一下，我在Windows 10（带有g++ 8.1）和Ubuntu 18.04 LTS（带有g++ 9.1）上都编译了程序。两者都打印1。