some friend functions don't follow the rule

Question

For the following snippet:

class A{
    friend void f(){};
    public:
        A(){f();} //error
};

class B{
    friend void f(void* ptr){};
    public:
        B(){f(this);} //no error
};

According to the rule that although friend functions can be defined inside a class, yet they are not visible until they are declared somewhere outside the class scope, the error in the definition of class A is explained.
But I am confused why the snippet for class B doesn't produce the same error as class A's.

Please can anyone tell me about this?

Putting `template` in front of the method seems to work as well. Leaves us with an ugly `f();` call though. — Caramiriel, Feb 25 '15 at 14:28
@Caramiriel That's a bug in GCC, won't work for other compilers, and won't work for later GCC versions once that bug is fixed. — , Feb 25 '15 at 18:08

Mike Seymour · Accepted Answer · 2015-02-25T23:02:15.760

"Not visible" is a bit of an over-simplification. With only an in-class definition, a friend function can't be found by qualified or unqualified lookup, which is why the first snippet fails.

However, it can be found by argument-dependent lookup (ADL), so you can call it with an argument involving a type that's scoped in the same namespace as the function.

In this case, the argument type is B*, scoped in the global namespace. The friend function is scoped in the namespace containing the class that declares it - also the global namespace. So ADL will look in the global namespace for functions called f, find the friend function, and use that.

some friend functions don't follow the rule

1 Answers1