Re: Correct mod/rem (was: Problem with % (remainder) operator

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Article #62

#62

From: Andy J Robb
Date: Mon, 22 Jan 1996 22:11

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hbaker@netcom.com (Henry Baker) wrote:

Personally, I agree - rounding towards minus infinity is preferred.
Generally, the IEEE seem to prefer bank rounding as the default.
This means that the value zero means:
	-0.5 <= answer <= 0.5
the value 1 means
	 0.5 <  answer <  1.5
the value 2 means
	 1.5 <= answer <= 2.5

Thus 3/-2 == -3/2 == -(3/2) == -(-3/-2) == -2
     2/-3 == -2/3 == -(2/3) == -(-2/-3) == -1

ISO/ANSI only guarantee that x/y*y + x%y = x (as you know).  The
remainder should be calculated after bank rounding the quotient.

Regards,

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Path: arc1-ca.pugleaf.net!s02.pavilion.co.uk!usenet
From: AJRobb@pavilion.co.uk (Andy J Robb)
Newsgroups: comp.lang.java,comp.lang.c,comp.lang.c++
Subject: Re: Correct mod/rem (was: Problem with % (remainder) operator
Date: Mon, 22 Jan 1996 22:11:18 GMT
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