Re: LZ: Variable bit?
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Re: LZ: Variable bit?
On Thu, 17 Jul 1997 Terry Peng <tpeng@geocities.com> writes:
>Help! I can't figure out what's wrong with this routine.
>Ok, let's say I need to access a variable bit in indirect memory.
>I have the variable in register b, except it's not your regular way of
>>numbering the bits.
>
> MSBit LSBit
>byte: 0 1 0 0 1 1 0 1
>regular way: 7 6 5 4 3 2 1 0
>my way: 1 2 3 4 5 6 7 0
>
>Would something like this work?
It looks to me like it would work. You can put the "jr z,endrotate"
before the "ld a,(de)" which will make it a little bit faster when
the bit number is 0. It is not necessary, because the "ld a,(de)"
doesn't change the condition codes.
>;input: b=no. of bit
>; (de)=byte
>;output: copy the bit to the carry flag
>GetBit:
> xor a ;same as ld a,0
> cp b ;b is the number of the bit
> ld a,(de) ;the byte to test is at (de)
> jr z,endrotate ;if b=0, goto endrotate
>rotate:
> rlca ;get the important bit into bit 0
> djnz rotate
>endrotate:
> srl a ;copy the LSBit to the carry flag
> ret
--
Patrick Davidson (ariwsi@juno.com)
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