Path: utzoo!utgpu!news-server.csri.toronto.edu!cs.utexas.edu!uunet!microsoft!jimad
From: jimad@microsoft.UUCP (Jim ADCOCK)
Newsgroups: comp.lang.c++
Subject: Re: Copying objects
Message-ID: <55658@microsoft.UUCP>
Date: 5 Jul 90 16:53:01 GMT
References: <18753@well.sf.ca.us> <4853@darkstar.ucsc.edu> <55612@microsoft.UUCP> <304@taumet.com>
Reply-To: jimad@microsoft.UUCP (Jim ADCOCK)
Distribution: comp
Organization: Microsoft Corp., Redmond WA
Lines: 24

In article <304@taumet.com> steve@taumet.UUCP (Stephen Clamage) writes:
|In article <55612@microsoft.UUCP> jimad@microsoft.UUCP (Jim ADCOCK) writes:
|>Interestingly, operator= has the unique characteristic of being
|>virtual but not inherited!  Can someone explain what this is suppose
|>to mean?
....
|The ARM states that the default operator= (memberwise assignment) is
|generated if (among other things) its address is taken.  If a derived
|class needs a virtual operator= and the programmer does not declare one,
|the compiler should generate the default version and put it in the
|virtual table.  If this didn't happen, it may be a bug in the compiler,
|or a compiler written to an earlier version of C++.

Look at the example again.  The compiler should only generate a default
operator= of the form D& D::operator=([const] D&).  But the virtual 
function whose template to be matched is B& D::operator=(const B&).
So presumably, a compiler should, in the special case of virtual 
operator=, give a compile time squawk if B& D::operator=(const B&)
isn't explicitly defined in every subclass D of B ?

[The compiler I tried this example on erroneously allowed D to inherit
B& B::operator=(const B&) via the vtable]

Still doesn't answer my question though -- what is this suppose to *mean* ???