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From: sdm@cs.brown.edu (Scott Meyers)
Newsgroups: comp.lang.c++
Subject: Re: Virtual function problem.
Keywords: C++ virtual function
Message-ID: <74942@brunix.UUCP>
Date: 8 May 91 18:47:03 GMT
References: <3472@trlluna.trl.oz> <718@taumet.com>
Sender: news@brunix.UUCP
Reply-To: sdm@cs.brown.edu (Scott Meyers)
Organization: Brown University Department of Computer Science
Lines: 37

In article <718@taumet.com> steve@taumet.com (Stephen Clamage) writes:
| 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);
| >};

...

|          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).

...

| 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*)

Nope, even when B::fn is virtual, it still hides the two versions of A::fn:
  	b.fn(2);	    // calls B::fn(int)
  	b.fn('c');	    // calls B::fn(int)
  	b.fn("hello");	// illegal
Remember that overloaded functions are disambiguated statically, and the
scope of B hides the scope of A.  Hence all references to the identifier fn
through an object of (static) type B will resolve to the fn in B's scope.

Scott


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