12 replies to this topic

[xian106]

I want to display a big graphics,but the right side is not the same as the left side.So I think it must be done using sprites.
And I think it is:

screen
---------------------------------------------------
p0 p1 p0 p1 p0 p1
p0 p1 p0 p1 p0 p1
...............
--------------------------------------------------

But when I write the second "p0",it will cover with the first "p0"
How to do it?
May be I must use "resp0" again before write the second "p0",but I can not do it....

[DEBRO]

Read the The scores / 48-pixel highres routine explained! thread on [stella]. This may help you.

[xian106]

Time is too hard to control....especially when the sprites moved.

[DEBRO]

xian106 said:

Time is too hard to control....especially when the sprites moved.

So what are trying to do? If it's a big moving sprite like in Dragster then read Eckhard's Movable 48 pixel sprite thread on [stella]. If you're looking for a formation like Space Invaders then read Paul Slocum's Atari 2600 Advanced Programming Guide. Read the section on the multi-sprite trick.

[vb_master]

Flicker?

[xian106]

OK,I get "bigmove.asm".I try to calculate its time,but have some trouble.
Could you have a look at:
[/code]

LDY TopDelay ;TopDelay =0 at the beginning
INY
;machine cycles
X1: STA Wsync ; 0,GO
DEY ; 2
BNE X1 ; 3--5
LDY #4 ; 2--7
X2: DEY ;
BPL X2 ; 5*4+5=25
; 25--32
LDA #9 ; 2 --34
STA LoopCount ; 3--37
JMP (DelayPTR) ;5--42 ;JMP TO $f241
JNDelay: byte $c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9
byte $c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9
byte $c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9
byte $c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9
byte $c9,$c5
; $f241 is here. C9=2 c5=3 5 --47
NOP ; 2 --49

X3: NOP ; 2 --51
NOP ; 2 --53
NOP ; 2 --55
LDY LoopCount ;3--58
LDA (s1),Y ; 5 --63
STA GRP0 ; 3 --66
LDA (s2),Y ; 5 --71
STA GRP1 ; 3 --74
LDA (s3),Y ; 5 --79(3)
STA GRP0 ; 3 --6
LDA (s6),Y ; 5 --11
STA Temp ; 3 --14
LDA (s5),Y ; 5 --19
TAX ; 2 --21
LDA (s4),Y ; 5 --26
LDY Temp ; 3 --29
STA GRP1 ; 3 --32
STX GRP0 ; 3 --35
STY GRP1 ; 3 --38
STA GRP0 ; 3 --41
DEC LoopCount ;5--46
BPL X3 ; 4 --50
; JMPX3 :NOP 2 --52


The problem is when p1 and p0 displayed?Could you label it behind the code?
In the X3 loop, there is 77 machine cycles?????

[xian106]

I am sorry. I donn't know how to modify.....I forgot the "[/code]"

         LDY     TopDelay          ;TopDelay =0 at the beginning 

         INY     

                ;machine cycles

X1:      STA     Wsync; 0    	

         DEY	; 2    	

         BNE     X1; 3--5    	

         LDY     #4 ; 2--7  	

X2:      DEY	;   

         BPL     X2; 5*4+5=25  	

          ; 25--32	

         LDA     #9; 2 --34	

         STA     LoopCount; 3--37  	

         JMP     (DelayPTR);5--42	;JMP TO $f241

JNDelay: byte    $c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9

         byte    $c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9

         byte    $c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9

         byte    $c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9,$c9

         byte  $c9,$c5; $f241 is here. C9=2 c5=3  5 --47	

         NOP  ; 2 --49	

X3:      NOP  ; 2 --51	

         NOP  ; 2 --53

         NOP  ; 2 --55

         LDY     LoopCount;3--58

         LDA     (s1),Y	; 5 --63

         STA     GRP0	; 3 --66

         LDA     (s2),Y	; 5 --71

         STA     GRP1	; 3 --74

         LDA     (s3),Y ; 5 --79(3)

         STA     GRP0	; 3 --6

         LDA     (s6),Y	; 5 --11;;;;;;;;;;;;;;;;;;;;;;;;

         STA     Temp	; 3 --14

         LDA     (s5),Y	; 5 --19

         TAX  ; 2 --21

         LDA     (s4),Y ; 5 --26

         LDY     Temp	; 3 --29

         STA     GRP1	; 3 --32

         STX     GRP0	; 3 --35

         STY     GRP1	; 3 --38

         STA     GRP0	; 3 --41

         DEC     LoopCount;5--46

         BPL     X3	; 4 --50

                	; JMPX3  :NOP  2 --52



                  

                  

        ;The problem is when p1 and p0 displayed?Could you label it behind the code?   

        ; In the X3 loop, there is 77 machine cycles?????	

[/code]

[Thomas Jentzsch]

You should follow the 1st link.

[xian106]

I had understood this problem.That's real hard.Time is important.
But there is a problem that I don not understand.It's about the pattern B position (RESP0) in the following program.I just use this program,but donn't undertand.

LDX	#$00;x =0,1,2,3,4 pattern =b c d,e f

	LDA	#30

	JSR	SetPatPos

----------------------------------------------------------------	

SetPatPos

	SEC

	STA	wsync	

.1	SBC	#$0F;2

	BCS	.1;3

	EOR	#$07;2

	ASL	;2

	ASL	;2

	ASL	;2

	ASL	;2        14+(a/$0f+1)*5 machine cycles

	STA	hmp0,X;4	B  C  D  E or F fine position

	STA	resp0,X;	pattern B,C,D,E,F position

	RTS

I had read this problem for a whole day~~
I donn't know why the value A-3 will be the actual position. Like:
A=#30 In this program,between the "STA WSYCN" and "STA RESP0,W" will take 14+(#30/$0f+1)*5=29 machine cycles, and HMP0=60(Shift left 6 pixels),So the B's position is 29*3-6=81 and 81-68=13 pixels (On TV).But I try it in games, the B's position will be 27 pixels(on TV) which is equal to A-3

A=#50，102-1=101 101-68=33 But it's 47 actually
A=#51，102-0=102 102-68=34 But it's 48 actually
A=#30， 87-6= 81 81-68=13 But it's 27 actually
A=#3 57-3= 54 54-68=-14 But it's 0 actually
Why it is?
I tried my best to understand it,but lost. I'm in China,and no friends could tell me.Writing games by oneself is not popular in China,and just programmer does it.I don't realize them. And no Atari program network for chinese on the Internet. So I ask help in USA,I think you could help me.
Thanks.
Don't laugh at me asking so fatuous problem I just study it by muself on the internet,and I like it very much.


The attachment is my try using this program.
I want to display "progra" at the middle of screen.But it has something wrong.I calculated its time carefully.I don't know why it is.Is it following season 1 or 2?
1 The reason is that I got a wrong position using above program. Maybe the position is not A-3....What number is it?.......
2 The reason is that time is wrong in my "display" loop.Please help me to check it.
Thank you.

Attached Files

•  progra.zip   1.99K   60 downloads

[Thomas Jentzsch]

Welcome to Atari 2600 programming Xian!

Ok, here is the code again, now using my usual labeling and with corrected cycle counting:

SetPatPos

  sec

  sta	WSYNC	

.1

  sbc   #$0F   ;2

  bcs   .1     ;2/3

  eor   #$07   ;2

  asl          ;2

  asl          ;2

  asl          ;2

  asl          ;2

  sta	HMP0,x ;4	

  sta	RESP0,x;4 = 18+n*5-1

  rts

So with a value of 30, hitting RESP0 happens at 18+3*5-1 = 32 (not 29).

The code above still has some minor offset (e.g. using 0, will not result in a displayinmg it at pixel 0), because with n=0, RESP0 happens at 22, which is IIRC, one cycle to early. But, unless the branch crosses a 256 byte boundary it should position the sprites pretty reliable.

[DEBRO]

xian106 said:

The attachment is my try using this program.
I want to display "progra" at the middle of screen.But it has something wrong.I calculated its time carefully.I don't know why it is.Is it following season  1 or 2?

Have you subscribed to [stella]? There are a lot of people there that can help with 2600 programming. It's a small community of VCS developers.

The reason why you're not seeing the word "progra" is because your routine has a bug. In your code you have...

    lda pdata_g,x    ;4--25

    tax        ;2--27

    ldy pdata_r,x    ;4--31

So you're overwriting your index with the sprite data.

[xian106]

DEBRO, Thank you for your help.I will have a try to modify my program.
And I have subscribed to stella just now.I think people in the Stella are proficient at 2600 program,it seems to be not wright that I ask them so simply question...

Thomas Jentzsch,Thank you.
"18+n*5-1" I think you mean I must add the 4 cycles with "sta RESP0,x",but why is a "-1" needed? Is it because the "bcs .1" ? When the C flag is set," bcs .1" just be 2 cycles?
OK,if it is "18+n*5-1",When A=3 , (18+1*5-1)*3-3-68= -5,but I had a try in games,when A=3, sprite will be displayed on the left side of screen,it means that the the position must be zero,and our result is -5.Why?

[Thomas Jentzsch]

xian106 said:

"18+n*5-1" I think you mean I must add the 4 cycles with "sta RESP0,x"

Yup, the reset gets triggered at the very last cycle of that instruction.

Quote

...,but why is a "-1"  needed? Is it because the "bcs .1" ? When the C flag is set," bcs .1" just be 2 cycles?

The other way arround: When C is clear (= branch not taken), the instruction requires only 2 cycles. So, when exiting the loop, we save one cylce (the "-1").

Quote

OK,if it is "18+n*5-1",When A=3 , (18+1*5-1)*3-3-68= -5,but I had a try in games,when A=3, sprite will be displayed on the left side of screen,it means that the the position must be zero,and our result is -5.Why?

I am not sure, if I understand you correctly here. I suppose ther eare two problems overlapping, making it hard to fix them.

1. problem:
A write to RESPx before cycle 23 works like a write at cycle 23. This is happening in your example above. Replace the last instruction with sta.w RESP0,x (= 5 cycles instead of 4).

2. problem:
The code above may not be 100% perfect. The might be a small offset (less than 3 pixel). Just add that offset at the start of the code and everything should be fine.
Note: missiles and IIRC the ball, have a slightly different offset than the sprites.

BTW: Don't be afraid to ask at [stella]. Just make sure that you check the archives before asking. If anything is still unclear (and this code is quite tricky), just ask, please.