Question

[Graziano Servizi]

I read in the 3.4.1.9 clause of the C++11 standard:

Name lookup for a name used in the definition of a friend function 
(11.3) defined inline in the class granting
friendship shall proceed as described for lookup in member function 
definitions. If the friend function is
not defined in the class granting friendship, name lookup in the friend 
function definition shall proceed as
described for lookup in namespace member function definitions.

Later, in the 11.3.6 clause I found:


A function can be defined in a friend declaration of a class if and only 
if the class is a non-local class (9.8),
the function name is unqualified, and the function has namespace scope. 
[ Example:
class M {
friend void f() { }
// definition of global f, a friend of M,
// not the definition of a member function
};
— end example ]


So, if I have well understood I could write such a short code:

# include <iostream>

class P
  {
   friend void friend_inline()
    {std :: cout << P :: Name << ' ' << p << '\n';}
   static constexpr int Name = 20;
   static constexpr double p = 12.5;

   };

int main()
  {
   P p;
   // how can I call friend_inline ?
}

which indeed is compiled by 4.8.3 g++ compiler. But the question is in 
the comment: I tried both the call as if friend_inline was in global 
scope (the standard said "global definition") and the call through the 
::  scope resolution operator such as

P :: friend_inline();

Neither worked. I also tried adding a public member to P as a "wrapper" 
of friend_inline (so making it and its friendship redundant), but the 
compiler claimed that friend_inline was undeclared in the class scope...

I can't understand...


                          Graziano Servizi