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From: beard@ux1.lbl.gov (Patrick C Beard)
Newsgroups: comp.lang.c++
Subject: Re: overloading ->
Message-ID: <4559@helios.ee.lbl.gov>
Date: 4 Jan 90 18:34:34 GMT
References: <10159@saturn.ucsc.edu>
Sender: usenet@helios.ee.lbl.gov
Reply-To: beard@ux1.lbl.gov (Patrick C Beard)
Organization: Lawrence Berkeley Laboratory, Berkeley
Lines: 59
X-Local-Date: 4 Jan 90 10:34:34 PST

In article <10159@saturn.ucsc.edu> daniel@spica.ucsc.edu (Daniel Edelson) writes:
#I have a container class for a pointer type.
#Is there a way to make -> work correctly
#w/o explicit coercion or using (*p).member?
#
#Quick example:
#
#struct object {
#	int a;
#};
#
#typedef object * Pobject;
#
#struct container {
#	Pobject p;
#	container(Pobject pp) : p(pp) { }
#	operator Pobject() { return p; }
#	object & operator * () { return *p; }
#};
#
#main()
#{
#	object o;
#	container p = &o;
#
#	(*p).a			// works but unnatural style
#	Pobject(p)->a	// works but explicit coercion is ugly
#	p->a;			// doesn't work
#}

The reason p->a doesn't work is that p is not a pointer to an object and
you haven't defined a member function operator->() for class container. 

Define:
Pobject container::operator->() { return p; }

#
#I would the user-defined conversion to take place
#before the -> operator is applied, but that doesn't appear
#to be what the language specifies. I've tried 3 distinct
#compilers (cfront 2.0, oregon, g++) and they all behave the
#same.
#

It doesn't make sense for the conversion to be done since, in your example,
you aren't dereferencing "p" when you write p->; p isn't a pointer.  So the
compilers are doing the right thing.

#Help is greatly appreciated,
#
#Daniel Edelson
#
#daniel@cis.ucsc.edu


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-  Patrick Beard, Macintosh Programmer                        (beard@lbl.gov) -
-  Berkeley Systems, Inc.  ".......<dead air>.......Good day!" - Paul Harvey  -
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