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From: steve@taumet.com (Stephen Clamage)
Newsgroups: comp.lang.c++
Subject: Re: Virtual function problem.
Keywords: C++ virtual function
Message-ID: <718@taumet.com>
Date: 8 May 91 16:09:25 GMT
References: <3472@trlluna.trl.oz>
Organization: Taumetric Corporation, San Diego
Lines: 41

mcf@tardis.trl.OZ.AU (Michael Flower) writes:

>class A {
>	virtual void	fn(char);	// 1
>	virtual void	fn(char *);	// 2
>};
>class B : public A {
>	void	fn(int);
>};

>"t.c", line 7: warning:  B::fn() hides virtual A::fn()
>"t.c", line 7: warning:  B::fn() hides virtual A::fn()

>Of course B::fn() hides A::fn(). That is why I wrote it! After all A::fn() is
>virtual.
>If I change the order of lines 1 and 2 (in class A), the problem goes away.

This is a compiler bug, but not in the way you think.  You should always
get the warning.  B::fn(int) hides (not overrides) both A::fn(char) and
A::fn(char*) in that neither function is implicitly accessible from a B
object.  A virtual function in a derived class overrides (not hides) one
in a base class if the signatures are the same (and the return types
must match as well).

In what you show, B::fn(int) is not virtual.  So with your code we can
get the following:
	B b;
	b.fn(2);	// calls B::fn(int)
	b.fn('c');	// calls B::fn((int)'c'), not A::fn('c')
	b.fn("hello");	// illegal
But you could still do this:
	b.A::fn('c');		// calls A::fn(char)
	b.A::fn("hello");	// calls A::fn(char*)

If you declare B::f(int) explicitly virtual, you will get:
	b.fn(2);	// calls B::fn(int)
	b.fn('c');	// calls A::fn(char)
	b.fn("hello");	// calls A::fn(char*)
-- 

Steve Clamage, TauMetric Corp, steve@taumet.com