[A83] Re: xor (hl)

To: <assembly-83@lists.ticalc.org>
Subject: [A83] Re: xor (hl)
From: "Ronald Teune" <rtwolf@gmx.net>
Date: Mon, 15 Oct 2001 19:14:29 +0200
Reply-To: assembly-83@lists.ticalc.org


Crappy:
xor (hl) = push bc \ ld b,(hl) \ xor b \ pop bc
(if all of these commands exist...)

>Yes, but the part what I meant was that hl can contain 2 Bytes, so my real
>question is how does a xor 2 bytes ??? (only L?, h with a and l with a??)
>
>
>
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>>From: Peter-Martijn Kuipers <hyper@hysoft-automation.com>
>>Reply-To: assembly-83@lists.ticalc.org
>>To: assembly-83@lists.ticalc.org
>>Subject: [A83] Re: xor (hl)
>>Date: Wed, 10 Oct 2001 19:59:26 +0200 (CEST)
>>
>>
>>Yep, it does.
>>
>>Suppose a contains %00111010,
>>
>>hl contains $6523 and the byte at $6523 is %10010111
>>
>>then this would happen
>>
>>a:     %00111010
>>(hl):  %10010111    XOR
>>-----------------------
>>new a: %10101101
>>
>>
>>
>> >
>> > i suppose it xor's the byte at adress hl...
>> >
>> > >
>> > > I'm a bit confused, xor uses the a register, right?
>> > > Then if you do xor (hl), does it then xor h with a, and l with a???
>> > >
>> >
>> >
>> >
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