From:	sparhawk AT eunet DOT at (Gerhard Gruber)
Newsgroups:	comp.os.msdos.djgpp
Subject:	Re: 64k demo
Date:	Mon, 29 Jun 1998 22:14:58 GMT
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Destination: "Arthur" <arfa AT clara DOT net>
From: Gruber Gerhard
Group: comp.os.msdos.djgpp
Date: Mon, 29 Jun 1998 19:45:44 +0100:

>You don't seem to understand what I mean. There is a fundamental difference
>between arethmetic and logical shifting. Arethmetic shifting shifts the data
>(bytes, words or longs) but keeps the sign bit intact. Logical shifting
>treats the sign bit as any other bit in the byte/word/long.
>
>So for instance, aretmetically shifting left the value -2 would give the
>result -4; logically shifting left -2 would give 4.

I think this is wrong. shifting left -2 always gives -4 because the bit
pattern still has the highest bit set. in hex this would be -2 = 0xFFFE and a
shift would result in 0xFFFC which still is -4. Problems would occure if you
shift left a value like 0x8001 because then the bit that shows the sign is
moved out and the result would be 0x0002 instead of 0x8002.

>The shift commands which change different bits that I was referring to are
>commands like "shift with carry" and "shift with overflow."

>AFAIK, you can only do arethmetic shifting in C, like I said.

I wrote a small program that shows this.

#include <stdio.h>

int main(void)
{
   long i1 = 0x80000002;

   printf("\n\n\n\n\n\ni1: %u %d %04X\n", i1, i1, i1);

   i1 <<= 1;

   printf("i1: %u %d %04X\n", i1, i1, i1);

   i1 = -2;

   printf("i1: %u %d %04X\n", i1, i1, i1);

   i1 <<= 1;

   printf("i1: %u %d %04X\n", i1, i1, i1);

   return(0);
}

and the output of this is:
i1: 2147483650 -2147483646 80000002

i1: 4 4 0004                    <- this is were it goes wrong because the
middle column should read -2147483645 instead of 4.

i1: 4294967294 -2 FFFFFFFE

i1: 4294967292 -4 FFFFFFFC

--
Bye,
   Gerhard

email: sparhawk AT eunet DOT at
       g DOT gruber AT sis DOT co DOT at

Spelling corrections are appreciated.

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