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Forums > C64 Coding > Updating sprite X msb
2016-03-26 16:50
ChristopherJam

Registered: Aug 2004
Posts: 509
Updating sprite X msb

Slack bastard time. Recs for updating $d010 if I've got an eight bit x-position in my object table, but all my sprites are offset a constant number of pixels to the right? Current draft is this:
	clc
	lda mobX,x
	adc xoffset
	sta $d000,y
	lda #0
	adc #255
	eor #255
	eor $d010
	and hibit,x
	eor $d010
	sta $d010

(x is both object ID and sprite number (I'll change how I index mobX later), y is double the sprite number. hibit contains 2**i)
 
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2016-03-26 19:43
The Human Code Machine

Registered: Sep 2005
Posts: 99
It's faster to unroll the code:

lda mobX
clc
adc xoffset
sta $d000
ror xhi
lda mobX+1
clc
adc xoffset
sta $d002
ror xhi

and so on

lda xhi
sta $d010
2016-03-26 19:49
Color Bar

Registered: Oct 2012
Posts: 99
Quoting The Human Code Machine
It's faster to unroll the code:

lda mobX
clc
adc xoffset
sta $d000
ror xhi
lda mobX+1
clc
adc xoffset
sta $d002
ror xhi

and so on

lda xhi
sta $d010


Nice! Yes, and you can use zeropage for xhi :-)
2016-03-26 19:55
The Human Code Machine

Registered: Sep 2005
Posts: 99
Yes, but it's only one cycle faster.
2016-03-26 21:25
ChristopherJam

Registered: Aug 2004
Posts: 509
Quoting Color Bar
When you update all sprites in one loop, you may update $d010 as follows..


Ah yes, I should probably have mentioned that my use case is a multiplexer component; for the first eight sprites I will indeed shift the bits in one by one.
2016-03-27 09:46
Hoogo

Registered: Jun 2002
Posts: 76
I wonder if that multiplexer reuses the sprites in order, so that for every call y will increased by 2 and start with 0 again? Not sure if that can make a difference, just thinking...
I guess offset is variable and not available when mobX is generated?
And are you already in need of rastertime, or is it just about making this a little more pretty?
	clc
	lda mobX,x
	adc xoffset
	sta $d000,y
	lda $d010
	bcc clear
	ora hibit,y
	bcs store
clear	and hibit+1,y
store	sta $d010
hibit	!by 1, 255-1, 2, 255-2, 4, 255-4...
2016-03-27 10:26
Hein

Registered: Apr 2004
Posts: 770
8 pieces of code to jump to, all other stuff is done outside multiplexer. Y holds the current list index.

    lax (sprite_index_lo),y
    lda sprite_y,x
    sta $d001
    lda sprite_x,x
    sta $d000
    lda sprite_a,x
    sta screen+$3f8
    cmp sprite_x_msb,x	;$00 if cleared, $ff if set
    lda $d010
    and #%11111110
    bcs +
    ora #%00000001
+   sta $d010
    lda sprite_c,x
    sta $d027
2016-03-27 10:39
Color Bar

Registered: Oct 2012
Posts: 99
2016-03-27 11:59
Color Bar

Registered: Oct 2012
Posts: 99
Quoting Hoogo
I wonder if that multiplexer reuses the sprites in order, so that for every call y will increased by 2 and start with 0 again? Not sure if that can make a difference, just thinking...



If you can use a separate piece of code for each sprite, maybe you could do (for example for sprite 3)?
clc
lda mobX+3
adc xoffset
sta $d006
lda $d010
and #f7 ; clear first
bcc store
ora #$08
store
sta $d010
2016-03-27 14:23
Skate

Registered: Jul 2003
Posts: 458
Quote: Quoting Hoogo
I wonder if that multiplexer reuses the sprites in order, so that for every call y will increased by 2 and start with 0 again? Not sure if that can make a difference, just thinking...



If you can use a separate piece of code for each sprite, maybe you could do (for example for sprite 3)?
clc
lda mobX+3
adc xoffset
sta $d006
lda $d010
and #f7 ; clear first
bcc store
ora #$08
store
sta $d010


I remember writing a macro like this one, using sprite number as a parameter and generating these and/or values according to the number. this way code looks fine, things speed up, memory usage sucx, speed rocx. :)
2016-03-27 19:50
Color Bar

Registered: Oct 2012
Posts: 99
Quoting Color Bar

If you can use a separate piece of code for each sprite, maybe you could do (for example for sprite 3)?

Now I look more carefully I see that this is what Hein is doing ...
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