Path: utzoo!telly!attcan!uunet!tut.cis.ohio-state.edu!bloom-beacon!apple!claris!hearn
From: hearn@claris.com (Bob Hearn)
Newsgroups: gnu.g++
Subject: Re: Isn't sizeof an operator?
Message-ID: <9120@claris.com>
Date: 21 Mar 89 00:14:08 GMT
References: <2352@buengc.BU.EDU> <8903202240.AA24954@yahi>
Reply-To: hearn@claris.com (Bob Hearn)
Distribution: gnu
Organization: Claris Corporation, Mountain View CA
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