Path: utzoo!attcan!uunet!lll-winken!ames!haven!adm!smoke!gwyn
From: gwyn@smoke.BRL.MIL (Doug Gwyn )
Newsgroups: comp.lang.c
Subject: Re: Broken compilers? (was Re: Comma Operator)
Message-ID: <9382@smoke.BRL.MIL>
Date: 15 Jan 89 00:07:40 GMT
References: <922@quintus.UUCP> <1904@buengc.BU.EDU>
Reply-To: gwyn@brl.arpa (Doug Gwyn (VLD/VMB) <gwyn>)
Organization: Ballistic Research Lab (BRL), APG, MD.
Lines: 21

In article <1904@buengc.BU.EDU> bph@buengc.bu.edu (Blair P. Houghton) writes:
->	printf("%d %d %d\n", (x = 1, y = 2), x, y);
>A while back, someone indicated that they knew of a compiler optimizer
>that would reduce something such as the above (ostensibly through
>constant-reduction) to
>printf("%d %d %d\n", (1,2), x, y);
>or maybe even
>printf("%d %d %d\n", 2, x, y);
>So, you compiler-writing C-programmers, is this thing broken or what?

The second reduction is of course equivalent to the first.
However, both are broken (assuming x or y is used afterwards),
because after the statement containing printf(), x is required
to contain the value 1 and y must contain the value 2.  Leaving
their contents unchanged is not permitted (unless the program
cannot detect it, e.g. if their values are afterwards unused).

>I mean, the assignment itself is not an expression per se, but a statement,

No, assignments are expressions, not statements.  Sticking a semicolon
after an expression turns it into a statement.