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From: davidsen@steinmetz.UUCP (Davidsen)
Newsgroups: net.lang.c,net.bugs
Subject: Re: e1?(void_e2):(void_e3)
Message-ID: <895@kbsvax.steinmetz.UUCP>
Date: Fri, 15-Aug-86 12:12:07 EDT
Article-I.D.: kbsvax.895
Posted: Fri Aug 15 12:12:07 1986
Date-Received: Sun, 17-Aug-86 09:12:48 EDT
References: <273@watmath.UUCP> <5858@alice.uUCp> <134@sas.UUCP> <155@cbnap.UUCP> <499@dg_rtp.UUCP> <1701@mmintl.UUCP>
Reply-To: davidsen@kbsvax.UUCP (Davidsen)
Organization: General Electric CRD, Schenectady, NY
Lines: 33
Xref: watmath net.lang.c:10266 net.bugs:875

In article <1701@mmintl.UUCP> franka@mmintl.UUCP (Frank Adams) writes:
>The other paradigm is that e1?e2:e3 is precisely one of e2 or e3, depending
>on the value of e1.  This is a very reasonable interpretation; but if it
>were correct, there would be one other important consequence which is not in
>fact legal.  This is that when e2 and e3 are lvalues, the compound
>expression should also be an lvalue.  In particular, one could write
>
>e1?e2:e3 = e4;
>
>which would mean the same thing as
>
>if (e1) then e2 = e4; else e3 = e4;
>

This kicked off an interesting thought:
  *(e1 ? <ptr expr> : <ptr expr>) = expr;

Lo and behold it does what the quoted expression indicates. In an
actual example:
  *(a < b : &b : &a) = 70;

I'm not sure it *good* for anything, but if I do it with macros using
cute names, I can enter it in the obfuscated C contest...
-- 
	-bill davidsen

  ihnp4!seismo!rochester!steinmetz!--\
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"Stupidity, like virtue, is its own reward"