Finding LSB in a register

[Kweepa]

Mine is 13 bytes (checked with www.asm80.com, which is awesome!).
Metalbrain's is 12 bytes, if you remove the ret c and replace ld hl,0 with xor h/xor l.

[djnzx48]

Could you replace XOR 255 with CPL and get it down to 11 bytes?

[Kweepa]

Yep, that works
Didn't know about that instruction!

[djnzx48]

If you're doing regular AND-then-OR masking, you can just use OR B instead of CPL: AND B (and the same for C) as it doesn't matter what the mask is when the pixel value is set. That would get it down to 9 bytes.

(EDIT: Actually in that case all you need to create the mask is a single DEC instruction. So this method only requires one byte.)

Maybe other types of masking would work, for instance OR for the mask and then XOR for the sprite. Are you doing the same thing to create a mask on the left using the MSB?

[Metalbrain]

and replace ld hl,0 with xor h/xor l.

xor h/xor l wouldn't set hl to 0, they would just make A = A xor H xor L

[arkannoyed]

Wow, some great ideas, and it certainly looks as if the 8 bit logic methods will win. I'll have a go with some ideas later and see what works best.

Lots of re-designing to do to make the printing byte-wise work as opposed to bit-wise, which is how the Chess pieces are done currently. The speed improvement will be huge though, and hopefully might lose a few bytes as well as a bonus!

[lister_of_smeg]

No loops:
ld bc,hl
dec bc
ld a,h
xor 255
and b
ld b,a
ld a,l
xor 255
and c
ld c,a
Just uses HL & BC

ld bc,hl isn't a valid z80 instruction

[arkannoyed]

yes, but ld B,H and ld C,L are, which does the same thing.

[arkannoyed]

Seems to almost do what it needs to. The only thing left that I should've mentioned before is that it needs to RESet the LSB in HL (the data register).
So whereas maybe HL=05c0h, after creating the mask, bit 6 of L needs to be cleared, giving the result HL=0580h.

Does that make sense?

[Kweepa]

yes, but ld B,H and ld C,L are, which does the same thing.

Wow, apparently asm80.com is too clever for its own good - it automatically converted ld bc,hl to ld b,h/ld c,l!

[lister_of_smeg]

Seems to almost do what it needs to. The only thing left that I should've mentioned before is that it needs to RESet the LSB in HL (the data register).
So whereas maybe HL=05c0h, after creating the mask, bit 6 of L needs to be cleared, giving the result HL=0580h.

Does that make sense?

This any good?

Code: Select all

ld	d, h
ld	e, l
dec	de
ld	a, d
xor	h
ld	d, a
cpl
and	h
ld	h, a
ld	a, e
xor	l
ld	e, a
cpl
and	l
ld	l, a
rr	d
rr	e