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From: scott@bbxsda.UUCP (Scott Amspoker)
Newsgroups: comp.lang.c
Subject: Re: % operator with negatives
Message-ID: <1529@bbxsda.UUCP>
Date: 14 Dec 90 22:57:42 GMT
References: <1990Dec12.185714.7169@mp.cs.niu.edu> <1990Dec12.205416.26622@zoo.toronto.edu> <7461:Dec1310:04:0990@kramden.acf.nyu.edu>
Reply-To: scott@bbxsda.UUCP (Scott Amspoker)
Organization: Basis International,  Albuquerque, NM
Lines: 20

In article <7461:Dec1310:04:0990@kramden.acf.nyu.edu> brnstnd@kramden.acf.nyu.edu (Dan Bernstein) writes:
>And if you're asking ``Where is the information on it?'', the answer is
>that a % b is defined so that (a / b) * b + (a % b) equals a. It's the
>ambiguous definition of / that makes % so useless with negatives.

Well I can really stir up the muck by pointing out that, by definition,

    a mod 0 == a

Therefore, the method of dividing (a/b) does not always work since b could 
legally equal 0.  To be fair, C compilers disregard this "feature" of a
modulo and give an error on expressions such as 'a % 0'.  If you wish
a guaranteed proper modulo you might have to implement your own and
not rely on the '%' operater.

-- 
Scott Amspoker                       |
Basis International, Albuquerque, NM | "I'm going out for a sandwich"
(505) 345-5232                       |                       - Ben
unmvax.cs.unm.edu!bbx!bbxsda!scott   |