Re: A83: HL\2

To: Olle Hedman <assembly-83@lists.ticalc.org>
Subject: Re: A83: HL\2
From: Linus Akesson <lairfight@softhome.net>
Date: Mon, 31 Aug 1998 19:26:37 +0100
Delivered-To: assembly-83-outgoing@towerguard.unix.edu.sollentuna.se
In-Reply-To: <3.0.3.32.19980831080838.006dfc50@hem.passagen.se>
Reply-To: assembly-83@lists.ticalc.org


On 31-Aug-98, Olle Hedman wrote:

>At 06:39 1998-08-31 +0900, you wrote:
>>
>>> Shift h right one step, then rotate l through carry to the right.
>>> 
>>> sra h
>>> rr  l
>>
>>Woah, could you explain that a little.  So the sra rotates h right one
>>step...what does the rr l do?  Thanks.

>look at it like this:
>contents in HL before:
>H:       L:
>10010101 01011100

>sra shifts H right one step like this:

>a zero is put in here ->01001010 <-the bit that was "pushed out" is put in
>Carry (important)

>rr rotates L right _through_carry_ like this:

>Here is the carry-bit put in (the bit that was pushed out from H)->10101110
>th bit pushed out is put in carry here to..  but we wont use that..

>so after, HL will look like this:
>H:       L:
>01001010 10101110

>thus shifted right one step, which is equal to have divided HL by 2..

>I hope that helped...
>//Olle

Thanks, that gave my keyboard a longer life. It sends its regards. =)

Linus

References:

Re: A83: HL\2

From: Olle Hedman <oh@hem.passagen.se>

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