[CSDb] - User Forums - problems w. opening sidebordes+sprites

You are not logged in - nap

CSDb User Forums

Forums > C64 Coding > problems w. opening sidebordes+sprites

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 08:11

Radiant

Registered: Sep 2004
Posts: 639

MagerValp: Mind = blown

2011-09-01 08:33

Frantic

Registered: Mar 2003
Posts: 1628

@Magervalp: Nice, yes! Stylish, yes! I think I have seen this kind of delay code before somewhere though. Am I right? Is it in some routine in C=Hacking or something?

Anyway.. I don't think your particular little routine is correct. :) Unless you call the routine with a JSR, there will be one RTS too much and you'll end up in stack space when RTS'ing to Eternia. ...but if you DO call the routine with a JSR it will waste a total of 44+6 cycles (with the initial JSR to the routine included) rather than 44 cycles. ...or did I miss something? Maybe this is actually how it was supposed to be?

Not as elegant, but unless I did some mistake, this routine should provide a relatively stylish wasting of 44 cycles (with the initial 6 cycles of the JSR to the waste44 routine included):

		[SOME CODE HERE]
		jsr waste44
		[SOME CODE HERE]


waste44:	;the routine is 9 bytes
		nop
		jsr :+
		jsr :+
:		nop
		rts

2011-09-01 09:58

WVL

Registered: Mar 2002
Posts: 886

check!

6 jsr delay44

2 nop
6 jsr 1
2 nop
6 rts

6 jsr 2
2 nop
6 rts

2 nop
6 rts

total : 44

2011-09-01 11:31

MagerValp

Registered: Dec 2001
Posts: 1059

Yes, it depends on if you count 6 cycles of the calling jsr or not. I couldn't be arsed to count the cycles of the previous examples :P

Don't know if I found it somewhere or if I thought of it myself.

2011-09-01 14:35

TWW

Registered: Jul 2009
Posts: 541

Dealy44:
    lda #%00010000
    sec
    ror
    // PAGEBREAK HERE
    bcc *-1
    rts  // A exits with #8 and done in 7 bytes.

Another variant with 1 byte less ;-)

2011-09-01 15:41

Oswald

Registered: Apr 2002
Posts: 5023

you are making it overcomplicated. 1 byte less and primitive:

$10=5

6 jsr delay44

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

2011-09-01 15:49

JackAsser

Registered: Jun 2002
Posts: 1989

Quote: you are making it overcomplicated. 1 byte less and primitive:

$10=5 6 jsr delay44 delay44 3 ldx $10 2 dex ; 3 bpl *-3 ;x5=25 cycles +when exits 4 cycles -> 29 6 rts

Yes, but you destroys teh regiztorz! == CHEAT! :D

2011-09-01 15:57

Oswald

Registered: Apr 2002
Posts: 5023

oh, didnt think about that :)

2011-09-01 18:13

TWW

Registered: Jul 2009
Posts: 541

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

Trust me I gave it some thought^^

2011-09-01 18:17

Mace

Registered: May 2002
Posts: 1799