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From: frank@zen.UUCP (Frank Wales)
Newsgroups: comp.arch
Subject: Re: Optimization vs. the programmer
Message-ID: <596@zen.UUCP>
Date: Thu, 23-Apr-87 22:07:57 EDT
Article-I.D.: zen.596
Posted: Thu Apr 23 22:07:57 1987
Date-Received: Sun, 26-Apr-87 04:41:59 EDT
References: <479@danews.ATT.COM> <3300003@uiucdcsm> <3135@jade.BERKELEY.EDU> <775@killer.UUCP>
Reply-To: frank@zen.UUCP (Frank Wales)
Organization: Zengrange Limited, Leeds, England
Lines: 57
Keywords: HP-41 HP-71 MOD
Summary: which mod?

In article <775@killer.UUCP> jfh@killer.UUCP (John Haugh) writes:
>Third grade math time ...
>	Try an example ...
>		23 / 8 = 2 r 7 right?
>	Now the tricky one ...
>		-23 / 8 = -2 r -7 <- my intuition says 'Yes'.
>
>This is the way I understand.  Do the expression (in integer math)
>
>	Remainder = Dividend - (Quotient * Divisor)
>
>So -
>	Remainder = -23 - (-2 * 8) = -23 - (-16) = -7.
>
>But how come when I ask my HP-41CV what it thinks, I get 1?  If I ask for
>INT of -23 / 8 I get -2, not -3. (Yes, -3 is floor (-23.0 / 8.0)).
...
>- John 		(jfh@killer.UUCP)

[this is a bit of a digression, but...]

The problem is that there is more than one way of computing the mod
function, and each has an effect on the outcome with negative numbers.
The HP-41 implements MOD as:

		MOD(Y,X) = Y - MINT(Y/X)*X	

where MINT(Y/X) is the maximum integer not larger than Y/X.
Special cases are where X==0 (result is Y) and Y==0 (result is 0). [1]

The HP-71 (the HP-41's big cousin) actually offers three different remainder
functions to get around this problem:

      MOD(X,Y), which works like the 41's MOD;
      RED(X,Y), which evaluates X - Y * N, where N is the nearest integer to
		X/Y -- it is the remainder function defined as REM in the IEEE
		Floating Point Standard;
      RMD(X,Y), which evaluates X - Y * IP(X/Y), where IP is the Integer Part
		of (X/Y) [as would be found by INT on the HP-41] -- it is
		the remainder function defined as REM in the ANSI BASIC
		Standard.
      
As implemented on the HP-71, X==Infinity or Y==0 yield an Invalid Arg error.[2]

Your mission, should you choose to accept it, is to decide which of these the
% operator in C represents (or even, which you think it should represent :-).

{
References:
[1] -- Extend Your HP-41 by W.A.C. Mier-Jedrzejowicz, Ph.D., p 53
[2] -- HP-71 Reference Manual, pp 190, 240, 252.
}

[digression ends...]

Frank Wales                 [frank@zen.uucp<->..!mcvax!ukc!zen.co.uk!frank]
Development Engineer, Zengrange Ltd, Greenfield Rd, Leeds, England, LS9 8DB.