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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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chatGPZ
Registered: Dec 2001
Posts: 11123 |
err, did you understand what the point of lfts post was? o_O |
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Zyron
Registered: Jan 2002
Posts: 2381 |
Obviously not. ;) |
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Flavioweb
Registered: Nov 2011
Posts: 447 |
Ok.
Forgot my previous post.
'Without access I/O regs'. |
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Slammer
Registered: Feb 2004
Posts: 416 |
Seams like the solution is combining Pexs hint with actual code and some kind of induction proof that ensures it always ends in the desired end state instead of an endless drift. |
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algorithm
Registered: May 2002
Posts: 702 |
I assume its probably using rmw instructions near/on badlines perhaps to remove jitter in stages? |
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Danzig
Registered: Jun 2002
Posts: 429 |
cli jsr *
worx ;) proof: One Year Crest |
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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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