Question

There is such code:

#include <iostream>

class A{

public:
    friend void fun(A a){std::cout << "Im here" << std::endl;}
    friend void fun2(){ std::cout << "Im here2" << std::endl; }
    friend void fun3();
};

void fun3(){
    std::cout << "Im here3" << std::endl;
}

int main() 
{  
    fun(A()); // works ok
    //fun2(); error: 'fun2' was not declared in this scope
    //A::fun2(); error: 'fun2' is not a member of 'A'
    fun3(); // works ok
} 

How to access function fun2()?

Answer 1

class A{

public:
    friend void fun(A a){std::cout << "Im here" << std::endl;}
    friend void fun2(){ std::cout << "Im here2" << std::endl; }
    friend void fun3();
};

Although your definition of fun2 does define a “global” function rather than a member, and makes it a friend of A at the same time, you are still missing a declaration of the same function in the global scope itself.

That means that no code in that scope has any idea that fun2 exists.

The same problem occurs for fun, except that Argument-Dependent Lookup can take over and find the function, because there is an argument of type A.

I recommend instead defining your functions in the usual manner:

class A {
   friend void fun(A a);
   friend void fun2();
   friend void fun3();
};

void fun(A a) { std::cout << "I'm here"  << std::endl; }
void fun2()   { std::cout << "I'm here2" << std::endl; }
void fun3();

Notice now that everything works (except fun3 because I never defined it).