Re: A83: Got another one for 'ya


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Re: A83: Got another one for 'ya




Or you could use:

      call  $+3
      nop
      nop
      pop   hl
      call  _ldHLind

(9 bytes and loads of clock cycles)

Linus

On 14-Jun-98, Florent Dhordain wrote:

>Hehe... This won't either works, as xor reg does A = A^REG ( the ^ means xor
>).
>So it only affect A.

>On the 83, you can only do xor a to clear a register. xor h will do A = A xor
>H,
>and won't clear anything.

>You can try :
>xor a \ ld h, a \ ld l, a  (stupid : 3 bytes, 12 clocks cycles )

>or :
>or a \ sbc hl, hl

>not so good : 3 bytes, 19 cycles...

>or use ld hl, 0 ( 3 bytes, 10 cycles ) ! This remains the fastest, incredible
>!

>- Florent

>-----Message d'origine-----
>De : Phelan Wolf <gwatford@worldnet.att.net>
>À : assembly-83@lists.ticalc.org <assembly-83@lists.ticalc.org>
>Date : dimanche 14 juin 1998 18:59
>Objet : Re: A83: Got another one for 'ya


>>
>>oops, xor h    xor l
>>only 2 clocks faster and 1 bytes smaller
>>
>>Olle Hedman wrote:
>>
>>> At 11:35 1998-06-14 -0700, Wolf wrote:
>>> >
>>> >1). instead of    ld hl,0    do    xor hl    it is 6 clocks faster and 2
>>> bytes smaller
>>>
>>> sorry, but you can't xor hl...
>>>
>>> //Olle
>>
>>
>>




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