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lft
Registered: Jul 2007
Posts: 369 |
Useless coding riddle: Stable raster without I/O
Hi!
I came up with a technique to synchronise code to the raster position without accessing any I/O registers. It is not very efficient, and hence not very useful, but it was a nice intellectual exercise.
This is the premise: Provide a small piece of code (less than a page) that may start executing at any time. When execution reaches the end of the code, the current rasterline and cycle will be known. You may assume that sprites and interrupts are off, and that d011 has its default value (9b).
See if you can figure out how it's done! |
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Slammer
Registered: Feb 2004
Posts: 416 |
My guess is that you could do it like this. Haven't tried it out, so there might be flaws...
Jevgr n ybbc jvgu rknpgyl bar ezj bcrengvba gung gnxrf gur rknpg nzbhag bs gvzr bs n senzr jura gur ezj bcrengvba fgrnyf na rkgen plpyr. Eha vg sbe rabhtu ahzore bs gvzrf gvy vg fgbcf va n fgnoyr pbaqvgvba. Jr abj xabj gur cbfvgvba bs gur ezj bcrengvba gb or bar bs 25 cbffvoyr cbfvgvbaf. Yrgf pnyy guvf cebprqher sbe 'Pnyvoengr'.
Abj nqq n qrynl sbe gur jubyr senzr zvahf 9 yvarf naq ercrng pnyvoengr ntnva. Jr ner abj ba bar bs gur svefg 24 cbffvoyr cbfvgvbaf. Ercrng guvf 25 gvzrf gb or fher gb or va cbfvgvba 1. Ceboyrz fbyirq.
Gb bcgvzvmr gur nobir lbh pbhyq bspnhfr fhogenpg zber guna avar yvarf. |
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Peiselulli
Registered: Oct 2006
Posts: 81 |
I think it can be done like Algorithm said before: Make a loop the fullfills the 19656 Cycles (PAL) for a frame with 25 stores to memory in it.
Program it in that way that this is only the case if these stores are just before of the bad lines. If this loop does not hit all 25 stores at right position, then is will last longer and will run over again until it is in sync.
But I'm too lazy to program it, because it is useless, sorry ... |
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Slammer
Registered: Feb 2004
Posts: 416 |
Peiselulli: Oh yeah and it will reach the stable condition faster. Didn't read Algos post like that though and you have to convince yourself that the possibly 25 extra cycles taken doesn't makes you miss the stable point, but I think you are right. |
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Fresh
Registered: Jan 2005
Posts: 101 |
Well, looks like I'm a bit late... anyway:
http://pastebin.com/PeP4BfU1
64 bytes long, probably shrinkable. |
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lft
Registered: Jul 2007
Posts: 369 |
@Freshness79:
It's great to see some actual code! This looks rather similar to my own solution (which happens to be 63 bytes). And the I/O registers d020 and d021 in your code can obviously be changed to addresses in RAM, for strict compliance.
I'm curious about one thing: In the large waiting loop (>7000 cycles) there's a write cycle. Did you make sure that it won't interfere with the synchronisation? |
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Fresh
Registered: Jan 2005
Posts: 101 |
Yes, of course I've put $d020 and $d021 just to show it works. But, now that I'm thinking about it, with a longer code I may be able to show it without touching I/O.
You're way smarter than me so I'm sure you know how, or maybe that's the way you've already implemented it. :)
About the write cycle (stx $d021), it shouldn't cause any trouble as it's misadligned compared to the INCs. |
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ruk
Registered: Jan 2012
Posts: 43 |
Boo! I thought that looping about $100 times would suffice, but no... 62 bytes though :)
https://gist.github.com/p-a/38abcb0091019743d55a |
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Fresh
Registered: Jan 2005
Posts: 101 |
Ok, this one does its job by using only two W cycles per frame. I think that's the minimum number, I doubt you can sync with one.
http://pastebin.com/HTuu4KPP
(48 bytes)
Beware, this is S*L*O*W!!! Use warp!
@ruk: hope you won't mind, I've borrowed something from your routine.
@lft: thanks for this nice challenge. And I'm still waiting a code that shows stability w/o I/O (ie no $d020 or such). |
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ruk
Registered: Jan 2012
Posts: 43 |
@Freshness, np, sharing is caring =)
Impressed by the size of your code! |
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lft
Registered: Jul 2007
Posts: 369 |
Here's my solution. I've squeezed it down a bit to 59 bytes, and it doesn't clobber S.
The code reaches "done" at the last rasterline, cycle 62, so one extra nop would bring you to the top of the screen.
I have verified that 385 iterations are enough through simulation. This gives a total execution time of about 7.7 seconds.
sei
lda #0
ldy #129 ; 385 = 256 + 129
loop0
ldx #169 ; 2
loop3
cmp (0,x) ; \_ 3210
cmp (0,x)
nop
dex
bne loop3 ; /
beq *+2 ; 3
and #1 ; 2
ora #48 ; 2
sec ; 2
loop1
jsr sub ; rrrwwr
sbc #2 ; 2
bcs loop1 ; 3/2
ldx #166 ; 2
loop4
cmp (0,x) ; \_ 3817
cmp (0,x)
cmp (0,x)
dex
bne loop4 ; /
cmp (0,x) ; 6
dey ; 2 \_ 11 (9 when leaving)
bne not0 ; 2/3
lsr ; 2
bcs done ; 2/3
back
jmp loop0 ; 3
not0
bne back ; 3 /
sub
ldx #89 ; 2
loop2
dex ; 2 \_ 444
bne loop2 ; 3/2 /
rts ; 6
done
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