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Krill
Registered: Apr 2002
Posts: 2980 |
Shortest code for stable raster timer setup
While working on my ICC 2019 4K entry (now postponed to ICC 2020, but i hope it'll be worth the wait), i came up with this (14 bytes):initstabilise lda $d012
ldx #10 ; 2
- dex ; (10 * 5) + 4
bpl - ; 54
nop ; 2
eor $d012 - $ff,x; 5 = 63
bne initstabilise; 7 = 70
[...]; timer setupThe idea is to loop until the same current raster line is read at the very beginning (first cycle) and at the very end (last cycle) of a raster line, implying 0 cycles jitter.
With 63 cycles per line on PAL, the delay between the reads must be 63 cycles (and not 62), reading $d012 at cycle 0 and cycle 63 of a video frame's last line (311), which is one cycle longer due to the vertical retrace.
The downside is that effectively only one line per video frame is attempted, so the loop may take a few frames to terminate, and the worst case is somewhere just beyond 1 second.
The upside is that it always comes out at the same X raster position AND raster line (0), plus it leaves with accu = 0 and X = $ff, which can be economically re-used for further init code.
Now, is there an even shorter approach, or at least a same-size solution without the possibly-long wait drawback? |
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Rastah Bar
Account closed
Registered: Oct 2012
Posts: 336 |
$08A3 |
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Jammer
Registered: Nov 2002
Posts: 1335 |
Quoting Rastah Bar$08A3
LOL! :D
So it's supposed to be in fixed place in order to work properly? I asked rather broad, relatively to usual structure of inits etc. |
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Rastah Bar
Account closed
Registered: Oct 2012
Posts: 336 |
I guess you did not follow the thread very carefully. The (currently) shortest code that can be placed anywhere was proposed in post #44. Quiss came up with a very bright idea in post #50 that uses the instabilities of the SHX instruction. It uses less RAM, but it has some restrictions on code location. Shorter variants were found, but they have much stronger location restrictions. |
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Copyfault
Registered: Dec 2001
Posts: 478 |
Quoting JammerStupid question from a layman - where this stabilizing piece of code is supposed to go exactly to do its job and not crash the whole thing? :) No worries, we're all laymen in some field(s);)
Short answer: it's a short routine that ends on a fixed cycle position of a PAL-rasterline.
If you want a bit more: quite often a raster stabilizing routine is needed for some funky raster code. One approach is to initialize a timer in such a way that you can read it at the beginning of your RASTER irq and use it as a counter for the cycle jitter. This requires some routine that actually initializes the timer somewhere in your init code.
Since every raster line has the exact same no. of cycles, it boils down to init the timer relative to that total no. of cycles of a line. Quiss came up with the initial idea to "wait for a badline" utilising the SHX abs,Y. So effectively his (and also my) routine loops until you hit a badline at an exact cycle position, thus ending up on a unique cycle after the badline. Here you'd usually insert your timer start trigger.
If you *really* want even more detail, feel free to pm me and I'll act as your personal explainer ;) |
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Jammer
Registered: Nov 2002
Posts: 1335 |
I understand stabilizing in genral but I wasn't sure if this one is supposed to be triggered onceat all, once per vbl, once per line or sth. To my knowledge timer based interrupts are supposed to be stabilized every call. That's at least what THCM does :) |
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Rastah Bar
Account closed
Registered: Oct 2012
Posts: 336 |
Quote: I understand stabilizing in genral but I wasn't sure if this one is supposed to be triggered onceat all, once per vbl, once per line or sth. To my knowledge timer based interrupts are supposed to be stabilized every call. That's at least what THCM does :)
The code examples in this thread are meant to run once prior to setting a timer at a precisely known cycle. This timer can then used everytime at the start of an interrupt routine to stabilize it (that is, compensate for the jitter). |
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Copyfault
Registered: Dec 2001
Posts: 478 |
Yes, Rastah Bar pointed it out already: all routines here are meant to run only once during init. Purpose of such a timer-init-routine is to know the exact cycle position (at least relative to a rasterline) at the end of the routine - without using timers, ofcourse;)
Main reason for adding yet another variant was Quiss' question wether his routine can be moved to mem-area $08xx. Turns out that it *is* possible;)
Ah, and to disarm the "routine must sit at a fixed mem-pos"-argument: with my approach the timer-init can be placed in almost any mem page - only the position within that page is fixed! So not fully flexible, but not too rigid either. |
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Copyfault
Registered: Dec 2001
Posts: 478 |
Found another one, but alas this time fully mem-adress-fixed:
$183c a0 9e ldy #$9e
$183e a2 19 ldx #$19
$1840 18 clc
$1841 10 fa bpl $083d
This saves another byte \o/
The CLC is just a 2-cycle place-holder that is replaced by $19 = ORA abs,Y when the SHX $19a2,Y hits the correct cycle in a badline. This ORA abs,Y effectively "eats up" the branch and ends the loop.
Mind we could also use other branch instructions (BCC springs to mind) but I decided to use BPL to avoid page-crossing (also saving a cpu cycle).
This example also shows that with SHX abs,Y and the likes, page-crossing must be carefully planned ($19a2 + y = $19a2 + $9e = $1a40, but the hi-byte is distorted s.t. it ends up as x & (hi+1) = $19 & $1a = $18). |
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Rastah Bar
Account closed
Registered: Oct 2012
Posts: 336 |
Quote: Found another one, but alas this time fully mem-adress-fixed:
$183c a0 9e ldy #$9e
$183e a2 19 ldx #$19
$1840 18 clc
$1841 10 fa bpl $083d
This saves another byte \o/
The CLC is just a 2-cycle place-holder that is replaced by $19 = ORA abs,Y when the SHX $19a2,Y hits the correct cycle in a badline. This ORA abs,Y effectively "eats up" the branch and ends the loop.
Mind we could also use other branch instructions (BCC springs to mind) but I decided to use BPL to avoid page-crossing (also saving a cpu cycle).
This example also shows that with SHX abs,Y and the likes, page-crossing must be carefully planned ($19a2 + y = $19a2 + $9e = $1a40, but the hi-byte is distorted s.t. it ends up as x & (hi+1) = $19 & $1a = $18).
The code location can't be $183c, can't it? Also the SHX instruction behaves unpredictable when a page is crossed, so I'm afraid this one won't work.
Btw, maybe you could summarize all the known allowed code locations where any SHX or SHY variant could work? |
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Copyfault
Registered: Dec 2001
Posts: 478 |
Quoting Rastah BarThe code location can't be $183c, can't it? Also the SHX instruction behaves unpredictable when a page is crossed, so I'm afraid this one won't work. Hm, I think it should, as the SHX has $19a2 as operand bytes. Adding Y=$9e and taking the wrong fixup into account gives $19a2 + $9e = $1a40 =(wrong fixup)= $1840. This is the adress of the CLC.
So starting the code at $183c is mandatory for that sniplet to work.
Quoting Rastah BarBtw, maybe you could summarize all the known allowed code locations where any SHX or SHY variant could work? I remember CJam had examined this in detail and posted it in some thread here on csdb. After a short forum scan I found it: https://csdb.dk/forums/?roomid=11&topicid=94460.
CJam summarised his findings in a very nice table (adjusted to SHX):high byte of address written to, when:
+--------+------------------+---------------+
| | no DMA on cycleN | DMA on cycleN |
+--------+------------------+---------------+
|page | | |
|not | H | H |
|crossed | | |
+--------+------------------+---------------+
|page | | |
|crossed | X&(H+1) | X&(H+1) |
| | | |
+--------+------------------+---------------+
value written, when:
+--------+------------------+---------------+
| | no DMA on cycleN | DMA on cycleN |
+--------+------------------+---------------+
|page | | |
|not | X&(H+1) | X |
|crossed | | |
+--------+------------------+---------------+
|page | | |
|crossed | X&(H+1) | X |
| | | |
+--------+------------------+---------------+
Here H is the hi-byte of the SHX-operand. For my example, this means there's page-crossing all the time (since y=$9e)- luckily, the strange hi-byte-fixup does not depend on the DMA-at-cylce-N-condition. |
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