From mboxrd@z Thu Jan  1 00:00:00 1970
To: Fans of the OS Plan 9 from Bell Labs <9fans@9fans.net>
In-reply-to: Your message of "Wed, 21 Apr 2010 15:40:19 +0200."
	<20100421134019.GA1382@polynum.com>
References: <20100421134019.GA1382@polynum.com>
From: Bakul Shah <bakul+plan9@bitblocks.com>
Date: Wed, 21 Apr 2010 16:28:52 -0700
Message-Id: <20100421232852.5AB865B4A@mail.bitblocks.com>
Subject: Re: [9fans] Integer arithmetic: some lessons
Topicbox-Message-UUID: 0b89a5e6-ead6-11e9-9d60-3106f5b1d025

On Wed, 21 Apr 2010 15:40:19 +0200 tlaronde@polynum.com  wrote:
> Hello,
>
> Still about integer arithmetic.
	...
> Conclusion (apparently): gcc always translate div involving power of two
> to binary manipulations, while (apparently) ken-cc does not.

gcc may choose to implement / with appropriate shifts and
what not but the result must be equivalent.  It certainly
can't replace %2 with >>1 for negative numbers.

Try this:

    int f(int x) { return x / 2; }
    int g(int x) { return x >> 1; }

    main(int argc, char** argv)
    {
	int x = argc > 1? atoi(argv[1]) : -7;
	printf("%d %d\n", -7/2, -7>>1);
	printf("%d %d\n", f(-7), g(-7));
	printf("%d %d\n", f(x), g(x));
    }

This will give you
    -3 -4
    -3 -4
    -3 -4
I don't have a way to test what kencc does but I would be
surprised if the result is any different....

>> is an arithmetic shift for signed numbers (the sign bit is
replicated).  The clearest way to see this is to compare

    -1/2 and -1>>1

The first gives you 0. The second gives you -1 (on twos
complement machines).

IIRC, Pascal's div operator behaves the same as C's / for
integers.  May be the problem is somehow related to mod?

> Conclusion: I will have to replace in METAFONT all div involving power
> of two to binary operations, since if I replace in some places and not
> in others, I wreak havoc the algorithms since computations are not done
> the same way for combined chunks.

Replacing div 2^N of a signed number by >>N seems like a mistake.