Path: utzoo!attcan!utgpu!jarvis.csri.toronto.edu!mailrus!purdue!bu-cs!buengc!bph
From: bph@buengc.BU.EDU (Blair P. Houghton)
Newsgroups: comp.lang.c
Subject: Re: powerful expressions
Message-ID: <5040@buengc.BU.EDU>
Date: 7 Dec 89 16:37:51 GMT
References: <12855@cit-vax.Caltech.Edu> <24735@cup.portal.com>
Reply-To: bph@buengc.bu.edu (Blair P. Houghton)
Followup-To: comp.lang.c
Distribution: usa
Organization: Boston Univ. Col. of Eng.
Lines: 29

In article <24735@cup.portal.com> Tim_N_Roberts@cup.portal.com writes:
>In <12855@cit-vax.Caltech.Edu>, wen-king@cit-vax.Caltech.Edu (King Su) shares:
>>	((qhead) ? (qtail = qtail->next = qnew)
>>		 : (qtail = qhead       = qnew))->next = 0;
>
>I am truly sorry for the order-of-evaluation flood that I am surely about
>to bring down upon myself, but...
>
>Is this guaranteed to work?  Does ANSI require a sequence point in
>(qtail = qtail->next = qnew) so that it works?  I have always been
>hesitant to write something like that, because I felt that the compiler
>could reasonably be expected to assign qnew to qtail first and thereby
>screw up qtail->next.  Am I fearing needlessly?

Fear no more.

The assignment operators group right-to-left.  That is

	qtail->next = qnew

is performed before

	qtail = qtail->next

is performed.

				--Blair
				  "K&R I, p. 191, for
				   what that's worth..."