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From: jfh@killer.UUCP
Newsgroups: comp.arch
Subject: Re: Optimization vs. the programmer
Message-ID: <775@killer.UUCP>
Date: Wed, 15-Apr-87 12:24:08 EST
Article-I.D.: killer.775
Posted: Wed Apr 15 12:24:08 1987
Date-Received: Sun, 19-Apr-87 03:17:40 EST
References: <479@danews.ATT.COM> <3300003@uiucdcsm> <3135@jade.BERKELEY.EDU>
Organization: The Unix(tm) Connection, Dallas, Texas
Lines: 46
Summary: Discussion of modulo and unsigned math ...

In article <3135@jade.BERKELEY.EDU>, desj@lemon.BERKELEY.EDU (David desJardins) writes:
> In article <3300003@uiucdcsm> grunwald@uiucdcsm.cs.uiuc.edu writes:
> >
> >Strength reduction is replacement of operations such as "i = j * 8" by
> >"i = j << 3", or "i = j % 8" by "i = j & 0x3" (if j is unsigned).
>                                   ^^^^^^^^^^^   ^^^^^^^^^^^^^^^^
>    I'm sure you mean "i = j & 0x7".  But, more importantly, this is the
> right remainder to compute even if j is signed (assuming two's-complement,
> of course).  It's too bad more people (especially hardware designers!)
> don't realize this.
> 
>    -- David desJardins

Okay - I have always assumed (No, not THAT word :-) that the code
	i = j & 0x7;
only worked with j being unsigned (or at least positive by use, e.g. array
subscripts).  My understanding is the % returns the remainder of j / 8.

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)).

New third grade math (binary example) ...
	Try an example ... (just remainders, no quotients)
		0010111 ^ 0000111 = 0000111 = 7 base 10. 
	Now the negative version ...
		1101001 ^ 0000111 = 0000001 = 1 base 10.

The first way seems to make sense - the second way works better but seems to
require a different quotient be calculated when performing the modulo.

- John 		(jfh@killer.UUCP)

No disclaimer.  Whattcha gonna do, sue me?