Path: utzoo!utgpu!news-server.csri.toronto.edu!mailrus!ames!decwrl!apple!agate!darkstar!terra.ucsc.edu!daniel
From: daniel@terra.ucsc.edu (Daniel Edelson)
Newsgroups: comp.lang.c++
Subject: scope of friend identifiers
Message-ID: <5708@darkstar.ucsc.edu>
Date: 3 Aug 90 17:45:52 GMT
Sender: usenet@darkstar.ucsc.edu
Reply-To: daniel@terra.ucsc.edu (Daniel Edelson)
Organization: University of California, Santa Cruz
Lines: 67

The rules governing names introduced in friend
declarations appear somewhat ambiguous according
to [E&S 90], section 11.4. It says 

	``If a class or function mentioned as a 
	friend has not been declared its name is 
	entered in the same scope as the name of 
	the class containing the declaration.''


Does this imply that a friend declaration in a nested
class declares a member function of the enclosing class?
E.g.,

	struct out {
		struct in {
			friend void frfunc();
		};
	};

According to the rule, the name frfunc is scoped to out.
Therefore, must I use out::frfunc to define the function?

	void out::frfunc() { /* ? */ }

Have I now declared a member function of out without using the
standard syntax? Or alternatively, is frfunc NOT a member of out,
but rather a function whose name is simply in the scope of out,
with scope and membership being somewhat orthogonal.

Or, on the contrary, am I interpreting the rule incorrectly.
Does it simple mean that the friend function is in the scope of out
in the following respect?

	struct out {
		static int outstat;
		struct in {
			friend void frfunc();
		};
	};

void frfunc()
{
	int i = outstat;  /* Am I in the scope of out? */
}

Or both?

Similar semantic ambiguities apply to class names first appearing
in friend declarations.

	struct out {
		struct in {
			friend struct frstruct;
		};
	};

Is struct frstruct a nested type of struct out?

Then again, in section 3.2 the discussion of class scope says that
``A name first declared by a friend declaration belongs to the 
global scope;'' which, if true, contradicts the 11.4 statement and
renders this discussion moot. Is there a diffinitive answer as to
which section is correct?

Daniel Edelson
daniel@cis.ucsc.edu