class friend function inside a namespace

c++, friend, namespaces

Solution

As well as adding a `::` you need to forward declare it, e.g.:

namespace A { class window; }

void f(A::window);

namespace A{
  class window{
    private:
    int a;
    friend void ::f(window);
  };
}

void f(A::window rhs){
  std::cout << rhs.a << std::endl;
}

Note that for this forward declaration to work you need to forward declare the class too!

Problem

Im trying to define a class friend function outside the namespace like this: ``` namespace A{ class window{ private: int a; friend void f(window); }; } void f(A::window rhs){ cout << rhs.a << endl; } ``` Im getting an error said that there is ambiguity. and there is two candidates `void A::f(A::window);` and `void f(A::window)`. So my question is : 1) How to make the global function `void f(A::window rhs)` a friend of the class A::window. EDIT: (After reading the answers) 2) why do I need to qualify the member function f inside window class to be global by doing `::f(window)` ? 3) why do I need to predeclare the function f(A::window) in this particular case, whereas when the class is not a defined inside a namespace it's okey for the function to be declared after the function is declared a friend.

Original source