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From: bright@dataio.UUCP (Walter Bright)
Newsgroups: net.lang.c,net.bugs
Subject: Re: e1?(void_e2):(void_e3)     so, is it legal or what?
Message-ID: <1061@dataio.UUCP>
Date: Wed, 13-Aug-86 12:29:57 EDT
Article-I.D.: dataio.1061
Posted: Wed Aug 13 12:29:57 1986
Date-Received: Thu, 14-Aug-86 22:01:14 EDT
References: <155@cbnap.UUCP> <499@dg_rtp.UUCP> <518@hadron.UUCP> <243@desint.UUCP>
Reply-To: bright@dataio.UUCP (Walter Bright
Organization: Data I/O Corp., Redmond WA
Lines: 46
Xref: mnetor net.lang.c:5635 net.bugs:319

In article <243@desint.UUCP> geoff@desint.UUCP (Geoff Kuenning) writes:
>While all of this discussion is very interesting and is important to
>achieving a precise language definition, let us not forget that there
>is *no* reason to ever write this particular expression.  Anywhere it's
>useful, you can just write
>
>	if (e1)
>	    void_e2;
>	else
>	    void_e3;

Not quite. This type of expression is very useful for macros which must
in themselves be expressions (to avoid peculiar {} and ; problems). Such
as:

#define biff(pow)	((pow) ? kayo() : oof())

Defining the macro as:

#define biff(pow)	if (pow) kayo(); else oof();

causes difficulties with constructs like:

	if (socko)
		biff(flez);
	else
		bang(glurp);

Similar problems exist for the other permutations of defining biff().

Now for the RIGHT (!) definition of when a void expressions is valid:

	A void expression is valid only under circumstances where the
	value of an expression is not used.

This means that:

	(a ? voidexp : voidexp),(a=b)

is valid. But a=a?voidexp:voidexp isn't valid, as the value is used. Apply
the rule above to all the cases and you have the RIGHT (!) answer. Note that
by implication, if the value of the ?: expression is not used, the operands
of the : need not be of compatible types, and need not be brought to a
common type.

I would like to see the ANSI C spec clarified on this point.