Path: utzoo!utgpu!jarvis.csri.toronto.edu!mailrus!tut.cis.ohio-state.edu!bloom-beacon!mit-eddie!WHEATIES.AI.MIT.EDU!gnulists
From: gnulists@WHEATIES.AI.MIT.EDU
Newsgroups: comp.lang.c++
Subject: Re: Isn't sizeof an operator?
Message-ID: <8903281246.AA26233@wheat-chex.ai.mit.edu>
Date: 28 Mar 89 12:46:28 GMT
Sender: daemon@eddie.MIT.EDU
Reply-To: hearn@claris.com
Lines: 36

In article <8903202240.AA24954@yahi> tiemann@lurch.stanford.edu writes:
>
>   In article <123@riunite.ACA.MCC.COM> rfg@riunite.UUCP (Ron Guilmette) writes:
>   ["why isn't sizeof declarable as an operator?"]
>
>   Because it's a compiler directive, not a C-language operator.  The
>   result of sizeof (something) is an integer _constant_.  If you try to
>   take the size of something that has a varying size, you will get only
>   the size of that thing at compile time.
>
>   If you're creating varying-sized data elements, you should keep track of
>   their sizes yourself.
>
>				   --Blair
>				     "But I don't know why you would..."
>
>It also takes a TYPE as an argument instead of an EXPRESSION.
>
>Michael

You are 100% incorrect.  Remember that the topic of discussion here is
C++, not C!  sizeof is certainly an operator.  And it can take a type
or an expression as an argument.  See Stroustrup, pp. 257-258.  It is,
however, resolved at compile time.  Presumably, it can't be overloaded
for this very reason; it is syntactically an operator, but it is not
treated the same as other operators.  It may be overloadable in the 
future; C++ is still an evolving language, and recently -> was added
to the list of overloadable operators.  However, it would be much nicer
if it used the dynamic rather than static information where appropriate,
like class/structure element dereferencing!!  Are you listening, Bjarne???

Unfortunately, for now, there seems to be no alternative but to keep track
of the sizes yourself.  Not very helpful, I know, but at least correct. :-)

Bob Hearn
hearn@claris.com