Path: utzoo!attcan!uunet!mcvax!hp4nl!orcenl!bengsig
From: bengsig@oracle.nl (Bjorn Engsig)
Newsgroups: comp.lang.c
Subject: Re: Adding two pointers
Message-ID: <325.nlhp3@oracle.nl>
Date: 11 May 89 12:07:12 GMT
References: <2765@buengc.BU.EDU> <563@lzaz.ATT.COM> <4093@ficc.uu.net> <922@garcon.cso.uiuc.edu> <924@garcon.cso.uiuc.edu> <113@ssp1.idca.tds.philips.nl> <1321@infmx.UUCP>
Reply-To: bengsig@oracle.nl (Bjorn Engsig)
Organization: ORACLE Europe, The Netherlands
Lines: 35

In article <1321@infmx.UUCP> kevinf@infmx.UUCP (Kevin Franden) writes:
=
=	struct{...}foo;
=
=	foo *p1,*p2,*p3;
/* I added the *'s in the above declarations */
/* And I would add the following: */
	int x,y;
=
=	x=37;
=	y=4;
=
=	p1=a[x];
=	p2=a[y];
=	p3=(x+y)*sizeof(foo);
=
= I realize that I missed part of this discussion, sorry if this has been
= posted already, but won't p3 be pointing to the 41st element?
Hmm, yes, but the construct is highly debatable ...
=                                                               if this
=is true, this is pointer addition, no?
No, it's integer addition, and a highly debatable computation of a pointer,
and an assignment which needs a cast.

In pointer arithmetic, you can achieve what you are trying by
	
	p3 = p2 + (p1 - a)

which means let p3 equeal the current p2 plus the number of elements between
p1 and the beginning of the array a.  Note, however that this is a pointer
plus an integer, the latter being the difference between two pointers.
-- 
Bjorn Engsig, ORACLE Europe         \ /    "Hofstadter's Law:  It always takes
Path:   mcvax!orcenl!bengsig         X      longer than you expect, even if you
Domain: bengsig@oracle.nl           / \     take into account Hofstadter's Law"