Path: utzoo!attcan!uunet!sdrc!scjones
From: scjones@sdrc.UUCP (Larry Jones)
Newsgroups: comp.std.c
Subject: Re: ANSI C token set (including $ and @)
Message-ID: <511@sdrc.UUCP>
Date: 20 Jan 89 22:59:39 GMT
References: <11343@haddock.ima.isc.com> <1858@zell.cs.vu.nl> <9438@smoke.BRL.MIL>
Organization: Structural Dynamics Research Corp., Cincinnati
Lines: 29

In article <9438@smoke.BRL.MIL>, gwyn@smoke.BRL.MIL (Doug Gwyn ) writes:
> In article <504@sdrc.UUCP> scjones@sdrc.UUCP (Larry Jones) writes:
> >That's what I thought, too.  But Karl pointed out to me that is
> >is possible to write a strictly conforming program that will NOT
> >be interpreted correctly by an implementation that allows '$' in
> >identifiers.
> 
> No, it isn't.  Use of the $ character in an identifier produces
> "undefined behavior".  The implementation of free to treat $ like
> _ in identifiers, because that cannot affect translation of any
> strictly conforming program.

But the critical point is that the $ character ISN'T in an
identifier if the implementation is conforming: foo$bar gets
parsed as three tokens just like foo+bar would.  As long as the $
doesn't make it past the preprocessor phases of translation, I
don't see anything in the standard that makes the program non-
conforming, and that makes any implementation that allows $ in
identifiers non-conforming since they do not parse the program
correctly and thus do not translate it correctly.

Please take another look at my (well, actaully Karl's) example.

----
Larry Jones                         UUCP: uunet!sdrc!scjones
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