Path: utzoo!attcan!uunet!ncrlnk!ncrcae!hubcap!gatech!ncar!mailrus!cornell!uw-beaver!microsoft!w-colinp
From: w-colinp@microsoft.UUCP (Colin Plumb)
Newsgroups: comp.lang.c
Subject: Re: Comma Operator
Message-ID: <267@microsoft.UUCP>
Date: 14 Jan 89 09:28:31 GMT
References: <922@quintus.UUCP>
Reply-To: w-colinp@microsoft.uucp (Colin Plumb)
Organization: very little
Lines: 33

nair@quintus () wrote:
> What should this print?
> 
> 	int x, y;
> 	printf("%d %d\n", (x = 1, y = 2), x, y);

(I assume you really want a third %d in there.)

It should print 2 <garbage> <garbage>.  If a machine evaluates arguments
left-to-right, it would print 2 1 2, but on many machines, arguments are
pushed right-to-left, and evaluated in the same order.  Thus, x and y
get assigned after their old values have been pushed.

C has never made any guarantees about the order in which arguments to
functions are evaluated; while the comma operator does, the comma between
function arguments is a different beast entirely.

> Shouldn't it be equivalent to:
> 
> 	int a, x, y;
> 	a = (x = 1, y = 2);
> 	printf("%d %d %d\n", a, x, y);

Ah, this is different... there's a sequence point at the ; after the
assignment to a, so it is guaranteed that x and y have their new values
before the call is made.

> Is there any justification in the first one printing
> 	2 1 0

Yes.
-- 
	-Colin (uunet!microsof!w-colinp)