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From: gintera@cs-sun-fsa.cpsc.ucalgary.ca (Andrew Ginter)
Newsgroups: comp.lang.c++
Subject: Overloaded a->b == (*a).b ?
Message-ID: <1991Mar25.225725.8625@cpsc.ucalgary.ca>
Date: 25 Mar 91 22:57:25 GMT
Organization: U. of Calgary Computer Science
Lines: 26
Originator: gintera@fsd
Nntp-Posting-Host: fsd


In section 13.4, ARM suggests:

  "For example, if more than one of the operators ->, *, and []
   are defined for a class V, one would expect v->m, (*v).m, and 
   v[-].m to have identical values for all objects v of class V."

At the same time, we find in section 13.4.6:

  "An expression x->m is interpreted as (x.operator->())->m for
   a class object x.  It follows that operator->() must either
   return a pointer to a class or an object of or a reference to
   a class for which operator->() is defined."

If operator-> returns a reference to a class for which operator -> is
defined, the processes of resolving the overloaded operator ->
continues recursively until something other than an object of or a
reference to a class with an operator-> is returned.  

This same is NOT true of an overloaded operator*().  It may therefore
be non-trivial for a programmer to maintain the a->b == (*a).b
equivalence.

Have I missed something here?

Andrew Ginter, 403-220-6320, gintera@cpsc.ucalgary.ca