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ChristopherJam
Registered: Aug 2004
Posts: 1409 |
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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ChristopherJam
Registered: Aug 2004
Posts: 1409 |
..and now
clc
lda mobX,x
adc xofse
sta $d000,y
lda $d010
ora hibit,x
bcs *+5
eor hibit,x
sta $d010
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BiGFooT
Registered: Mar 2002
Posts: 33 |
not tested, lazy solution below:
clc
lda mobx,x
adc xoffset
sta $d000,y
bcs set ; hibit doesn't count at overflow
lda hibit,y
beq clear
set
lda $d010
ora table,y
bne done
clear
lda table,y
eor #$ff
and $d010
done
sta $d010
table .byte 128,128,64,64,32,32,16,16,8,8,4,4,2,2,1,1 |
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ChristopherJam
Registered: Aug 2004
Posts: 1409 |
Good idea indexing hibit with y instead of x.
I'm not going to need to bash the sprite colours for present purposes, possibly not even sprite definition, so I may be able to avoid placing sprite number in a register altogether.
The rest looks sound, and looks like it sits between my OP and my reply to myself in speed.. |
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Rastah Bar
Account closed
Registered: Oct 2012
Posts: 336 |
When you update all sprites in one loop, you may update $d010 as follows
ldx #$07
loop:
ldy table,x ;y contains 2 times x
clc
lda mobX,x
adc xofse
sta $d000,y
rol $d010 ;shift in a 0 or 1
dex
bpl loop
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The Human Code Machine
Registered: Sep 2005
Posts: 112 |
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 |
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Rastah Bar
Account closed
Registered: Oct 2012
Posts: 336 |
Quoting The Human Code MachineIt'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 :-) |
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The Human Code Machine
Registered: Sep 2005
Posts: 112 |
Yes, but it's only one cycle faster. |
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ChristopherJam
Registered: Aug 2004
Posts: 1409 |
Quoting Color BarWhen 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. |
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Hoogo
Registered: Jun 2002
Posts: 105 |
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...
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Hein
Registered: Apr 2004
Posts: 954 |
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
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