Friends confusion

Question

$11.4/5 - "[...]A friend function defined in a class is in the (lexical) scope of the class in which it is defined[...]"

What does this statement mean?

struct A{
   typedef int MYINT;
   void f2(){f();}                    // Error, 'f' is undefined
   friend void f(){MYINT mi = 0;}     // Why does this work, shouldn' it be A::MYINT?
   void f1(){f();}                    // Error, 'f' is undefined
};

int main(){}

1. What is confusing here is that the call to 'f' from 'A::f1' is quiet understandable. However why is call to 'f' from 'A::f2' ill-formed, when a friend is in 'lexical' scope of the befriending class? What does 'lexical' scope mean?

2. At the same type why is the usage of 'MYINT' in 'f' OK? Shouldn't it be 'A::MYINT'?

If I add a parameter of type 'A *' to 'f', then both 'f1' and 'f2' are able to find 'f' because of ADL. This is understandable.

Answer 1

You have quoted only part of §11.4/5. According to it f() should be declared out of class first (function should have namespace scope). Try this:

void f(); // declare it first
struct A{
   typedef int MYINT;
   void f2(){f();}                    
   friend void f(){MYINT mi = 0;}     // definition of global f, a friend of A
   void f1(){f();}                    
};

As for second question, it is ok because of quoted by you part of §11.4/5. f() obeys the same rules for name binding as a static member function of that class and has no special access rights to members of an enclosing class.