[A83] Re: help for a routine

To: <assembly-83@lists.ticalc.org>
Subject: [A83] Re: help for a routine
From: "David Phillips" <david@acz.org>
Date: Wed, 18 Sep 2002 18:03:44 -0500
References: <IHUQWUCZWHBZU73LGLJC8TOXPNKFUT.3d85bc26@leshoffmann@9online> <000601c25f3c$21b54700$080a0a0a@ronald>
Reply-To: assembly-83@lists.ticalc.org

Ronald Teune writes:
> How would you do h*256? It's much more than 255 so I guess 0 comes
> out of it (??), unless you'd use a two byte register.

Not exactly.  You would perform the operation on a 16 bit register.  The
simplest way, as my example showed, is to use HL and simply set L to 0.  If
E is less than H then the result of the division will not fit into A.

--
David Phillips <david@acz.org>
http://david.acz.org/

References:

[A83] help for a routine

From: guillaume.h@ifrance.com

[A83] Re: help for a routine

From: "Ronald Teune" <rtwolf@gmx.net>

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