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From: tim@ism780c.UUCP (Tim Smith)
Newsgroups: net.arch
Subject: Re: Right shift vs. divide (change divide!)
Message-ID: <257@ism780c.UUCP>
Date: Wed, 15-Jan-86 16:00:31 EST
Article-I.D.: ism780c.257
Posted: Wed Jan 15 16:00:31 1986
Date-Received: Sat, 18-Jan-86 07:05:17 EST
References: <124000005@ima.UUCP> <4772@alice.UUCP> <1016@turtlevax.UUCP> <32@calgary.UUCP> <345@mcgill-vision.UUCP>
Reply-To: tim@ism780c.UUCP (Tim Smith)
Organization: Interactive Systems Corp., Santa Monica, CA
Lines: 26

In article <345@mcgill-vision.UUCP> mouse@mcgill-vision.UUCP (der Mouse) writes:

>  C (I can hear it now:  "Doesn't >> mean `much greater than'?").

I thought it was an input operator.. :-)

>     Only reason I can see for a  mathematician -- for anyone -- to want
>it  to   work  this  way  is   that  then  x-(y*(x/y))  is  always  >=0.

Actually, it is not that we want it to be >= 0, we want it to have only |y|
different values, rather than 2|y|-1 values. In fact, if y < 0, then we
want it to be <= 0.

> I don't see why you aren't satisfied with -3/2=-1 and -3%2=-1.
> You still have that x=(x%y)+(y*(x/y)), which is really all
> mathematicians demand, as far as I know (flame retardant: my
> knowledge of number theory is limited to undergrad level).

I want (x+ny)/y to be the same as x/y + n.  This only works if we
always round toward negative infinity, or always round toward positive
infinity.

>     What do you want done with -3/-2 or 3/-2?
1 and -2.
-- 
Tim Smith       sdcrdcf!ism780c!tim || ima!ism780!tim || ihnp4!cithep!tim