Path: utzoo!utgpu!watmath!clyde!att!osu-cis!tut.cis.ohio-state.edu!mailrus!ncar!umigw!steve
From: steve@umigw.MIAMI.EDU (steve emmerson)
Newsgroups: comp.lang.c
Subject: pointer comparison question
Summary: is this guaranteed or not?
Message-ID: <185@umigw.MIAMI.EDU>
Date: 14 Nov 88 00:58:26 GMT
Reply-To: steve@umigw.miami.edu (steve emmerson)
Followup-To: comp.lang.c
Distribution: na
Organization: University of Miami
Lines: 29

Sorry to bring this up again, but I didn't get no satisfaction last time
(i.e. no one responded to the main point in question).

The question is this: is the following expression guaranteed to return 
true if and only if "TestPtr" points within the array:

static struct foo	array[NUM];
static struct foo	BegPtr	= array[0];
static struct foo	BegPtr	= array[NUM-1];
extern struct foo	*TestPtr;				/* could also come from a
											   function call */

TestPtr >= BegPtr && TestPtr <= EndPtr;		/* guaranteed to do what
											   I want? */

Or is it "implementation defined" or "undefined".  Note the "if and only
if" condition, which means that if the expression is false, then "TestPtr" 
*must* point "outside" the array.

I know about pointer comparisons only being guaranteed for pointers into
the same array.  What I don't know is what that means for the above
expression.

P.S.  This is a contrived problem.  I know the alternatives.
-- 
Steve Emmerson                     Inet: steve@umigw.miami.edu [128.116.10.1]
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"Computers are like God in the Old Testament: lots of rules and no mercy"