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2011-08-30 22:02

Norrland

Registered: Aug 2011
Posts: 14

problems w. opening sidebordes+sprites

Hi there! First post here, but bare with me, I'm gonna spam you with some questions on semi-newbie level for a while..

I'll start with my problems with opening the sideborder. I'm showing a bitmap picture and the plan is to have 4 sprites in the sideborder, in the middle of the screen, all on the same y-pos. I have managed to make a stable irqroutine and have set $d015 to #$f0 to enable sprites 4-7 (sprites in order, the ones with lowest prio). I've also made sure that sprites 0-3 have totally different y-pos than sprites 4-7.
Later down the screen, at the same position as the sprites, I open the border with dec/inc $d016 (cycle 56), and everything goes well for the normal lines, but on bad lines I'll be 3 cycles late (if I've understood everything right) even though I do dec/inc $d016 right after the last lines dec/inc.
I've tried to follow Christian Bauers (+codebase, c-hacking, posts here and others) texts about the vic, rastertiming and opening borders, but I don't understand what I do wrong. In my code I'll do 20 nops between the dec/inc, which in my head equals to 52 cycles including dec/inc, and if that is right, suggests that the vic uses 11 cycles (63-52=11) (2/sprite and 3 for BA signal??) for fetching spritedata for 4 sprites on non-badlines.
If my calculations are right, I don't understand why I have problems on badlines, were I should have 23 cycles(?). Even if I skip one sprite, I'm still one cycle late, and I've read that 4 sprites should be possible..

The code inside irq, and screenshot showing my timing:

dec $d021 ;row1
inc $d021

.byte $ea, $ea, $ea, $ea, $ea, $ea, $ea ;20 st (40 cycles)
.byte $ea, $ea, $ea, $ea, $ea, $ea, $ea
.byte $ea, $ea, $ea, $ea, $ea, $ea

dec $d021 ;row2 6 cycles
inc $d021 ; 6 cycles,tot 52

.byte $ea, $ea, $ea, $ea, $ea, $ea, $ea ;20 st (40 cycles)
.byte $ea, $ea, $ea, $ea, $ea, $ea, $ea
.byte $ea, $ea, $ea, $ea, $ea, $ea

dec $d021 ;row3 6 cycles
inc $d021 ; 6 cycles,tot 52

dec $d020 ;row 4 BADLINE hell breaks loose... (nåja)
inc $d020

http://i1120.photobucket.com/albums/l491/lordborak/kastabort.png

Is my understanding right? Do I need to consider/setup anything else than described above?

... 25 posts hidden. Click here to view all posts....

2011-09-01 21:25

Slammer

Registered: Feb 2004
Posts: 416

:pause #$2c (From codebase, a bit down on the page)
Just put in your own optimizations.

2011-09-02 04:57

Oswald

Registered: Apr 2002
Posts: 5017

Quote: plus you need to set $10 which will set you back 4 bytes ;-)

Trust me I gave it some thought^^

yeah, thats why you destroy A, plus $10 doesnt needs to be set each time, so in the long run it uses less mem anyway :)

2011-09-02 05:43

Martin Piper

Registered: Nov 2007
Posts: 634

I've often thought some kind of external tool that generates ASM would be good here. You tell the tool what registers/memory you want set to what value and the raster and cycle it should be set on. Then it figures out the necessary most optimal timed code including bad lines, sprite DMAs etc.

2011-09-02 06:23

Perplex

Registered: Feb 2009
Posts: 254

If you write a tool in some high level language to generate the timed code for you, better make it waste cycles in between the timing critical instructions by interleaving it with other useful code instead of just nops and the like.

2011-09-02 06:37

JackAsser

Registered: Jun 2002
Posts: 1989

Quote: If you write a tool in some high level language to generate the timed code for you, better make it waste cycles in between the timing critical instructions by interleaving it with other useful code instead of just nops and the like.

That is exactly what the code multiplexer in S:T Lars Meeting III - Invite does. It has two code snippets, one that updates the 4x4-effect using A,X and Y registers. And one that opens the border. The multiplexer then replaces all NOPs in the timing critical code with the code from the 4x4-updater, keeping track of A,X and Y usage etc.

2011-09-02 11:47

TWW

Registered: Jul 2009
Posts: 541

Quote: yeah, thats why you destroy A, plus $10 doesnt needs to be set each time, so in the long run it uses less mem anyway :)

A was 8 upon entry and is 8 upon exit.

lda #$08 <----- see, 8!
ldy #$00
OPEN DA BOOORDEEEERRZZZZZZZ!!!!!
Still 8 here!

You however, destroy X and add bytes ;-)

Come to think of it, one could probably use X as a counter to shorten this even further...

<-entrypoint (X set with badline offset $02 set with quantities of linesX8)
!: jsr delayXX
sta $d016
sty $d016
dex
bne !-
jsr delayXX-1
sta $d016
sty $d016
sta $d016,x
sty $d016
ldx #$07
jsr delayZZ
dec $02
bne !-+3
rts

Then again this is straig out of my ass^^

2011-09-02 12:59

Oswald

Registered: Apr 2002
Posts: 5017

cool, then I win by simply modifying my code to:

$10=6

delay44
3 ldx $10
2 dex ;
3 bne *-3 ;x5=25 cycles +when exits 4 cycles -> 29
6 rts

x needs to be 0, and my x wil be 0 :)

edit: you also need to be on a pagebreak, which is not a nice thing either :)

2011-09-02 13:50

Frantic

Registered: Mar 2003
Posts: 1627

Quote: cool, then I win by simply modifying my code to:

$10=6

delay44
3 ldx $10
2 dex ;
3 bne *-3 ;x5=25 cycles +when exits 4 cycles -> 29
6 rts

x needs to be 0, and my x wil be 0 :)

edit: you also need to be on a pagebreak, which is not a nice thing either :)

...but is it stylish?

2011-09-02 15:36

Oswald

Registered: Apr 2002
Posts: 5017

it does achive what it was intended to: do it in less bytes. changing the goal later, isn't that unfair ? :)

2011-09-02 17:10

TWW

Registered: Jul 2009
Posts: 541

Fair enough. Pagebreak isn't cool, but a still somewhat if someone rips your code and don't think about it (hehe).

You still need to do: sta/y/x $10 "somewhere" to ensure $10 = 6. Thus adding 2 bytes (maybee 2 more unless you have a state of 6 in one of your regs) and yielding a total consumption of 8 bytes vs. my 7 bytes.

I guess you would use 3 cycles more overall due to the sta/sty/stx $10 aswell ;-D

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