A little help for a simple machine code program.

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peacer
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A little help for a simple machine code program.

Post by peacer » Mon Jul 29, 2013 6:16 pm

I am trying to write a simple machine code routine.

I want that program to scan hires screen memory to look for a color byte and change it to another.

I mean, for example, it scans for "1" value in HIRES memory and change all the "1"s to "4" . So in the picture, all the red parts turns to the blue.

Here's the code I wrote but it does not work. AgAin I am not good at assemble monitors or C. I write machine code programs under basic by entering hex codes. So I will give hex codes and try to explain what I ment to with equivalent BASIC tokens.

'DOKE 2,#A000 --> BEGINNING OF HIRES
$1000 A9 00 LDA #$00
$1002 85 02 STA $02
$1004 A9 A0 LDA #$A0
$1006 85 03 STA $03

'DOKE 4,#1F40 ----> Size of hires - 8000 bytes
$1008 A9 40 LDA #$40
$100A 85 04 STA $04
$100C A9 1F LDA #$1F
$100E 85 05 STA $05

'Y=0: A=PEEK(DEEK(2))+Y
$1010 A0 00 LDY #$00
$1012 B1 02 LDA ($02),Y

'IF A=PEEK(0) THEN POKE A,PEEK(1)
$1014 C5 00 CMP $00
$1016 D0 04 BNE $101C
$1018 A5 01 LDA $01
$101A 91 02 STA ($02),Y

'DOKE 2,DEEK(2)+1
$101C 18 CLC
$101D E6 02 INC $02
$101F D0 02 BNE $1023
$1021 E6 03 INC $03

'DOKE 4,DEEK (4)-1
$1023 C6 04 DEC $04
$1025 A5 04 LDA $04
$1027 C9 FF CMP #$FF
$1029 C6 05 DEC $05

'IF DEEK (4) > 0 THEN GOTO $1010
$102B D0 E3 BNE $1010
$102D 60 RTS


My program is like that. It is supposed to look for memory location 0 and change every byte in HIRES which is the same as PEEK(0) TO PEEK(1)

So I do this but nothing changes . It immediately says ready.
POKE 0,1:POKE 1,4 : CALL#1000

I might have errors in branchings and increasing and decreasing the memories.


Can you check and tell me what is wrong?

Or you might suggest easier way to search for a memory and change it like that.

Thank you very much .

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Chema
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Re: A little help for a simple machine code program.

Post by Chema » Mon Jul 29, 2013 7:25 pm

Okay, let me give it a very quick glance to see if I get something:

Code: Select all

'Y=0: A=PEEK(DEEK(2))+Y
$1010 A0 00 LDY #$00 
$1012 B1 02 LDA ($02),Y 
Mmm... you are not doing in MC what you type in BASIC. LDA($02),y would do something like A=PEEK(DEEK(2)+Y) instead. Is it what you want? I think it is.... Anyway Y is set to zero, so it is A=PEEK(DEEK(2)). Please forbid me if I get a bit confused with the parenthesis. It's been ages since I last programmed in BASIC...

Code: Select all

'IF A=PEEK(0) THEN POKE A,PEEK(1)
$1014 C5 00 CMP $00 
$1016 D0 04 BNE $101C 
$1018 A5 01 LDA $01 
$101A 91 02 STA ($02),Y 
The comparison is OK, I think. If the contents of register A and the contents of the address $00 are equal then you load 1 into A and store the contents of A (1) into DEEK(2)+Y... therefore something like POKE (DEEK(2)+Y),A or, as Y=0; POKE DEEK(2),A

Code: Select all

'DOKE 2,DEEK(2)+1
$101C 18 CLC 
$101D E6 02 INC $02 
$101F D0 02 BNE $1023 
$1021 E6 03 INC $03
Clearing the carry is not necessary, as you are not adding, subtracting rotating etc... The above code then increments a 16-bit value stored in $02,$03.

Code: Select all

'DOKE 4,DEEK (4)-1 
$1023 C6 04 DEC $04 
$1025 A5 04 LDA $04 
$1027 C9 FF CMP #$FF 
$1029 C6 05 DEC $05
Now you decrement a 16-bit value, but the code seems to be wrong as after the comparison CMP #$ff you should add a branch instruction, something like BNE $102b so you don't decrement the contents at $05.

If I am not wrong, the above code decrements both values at $05 and $04 all the time.

Code: Select all

'IF DEEK (4) > 0 THEN GOTO $1010 
$102B D0 E3 BNE $1010 
$102D 60 RTS 
This will only check if the Z flag is set after decrementing the contents of $05, not of the whole 16-bit value being zero. You can do that with something like LDA $04: ORA $05: BNE $1010

So basically I think that the problem is in the decrementing and the comparison, as you imagined.

Hope it helps...

Cheers.

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Re: A little help for a simple machine code program.

Post by peacer » Mon Jul 29, 2013 8:26 pm

:) Thank you very much for the quick response. I think I had errors in basic explanations .
$1010 A0 00 LDY #$00
$1012 B1 02 LDA ($02),Y

Yes, I wanted to set A to PEEK(DEEK(2)) . As there is no such command, I use indirect indexing with Y and setting Y as zero. As long as Y is zero, this part is fine then.
Code: Select all
$1014 C5 00 CMP $00
$1016 D0 04 BNE $101C
$1018 A5 01 LDA $01
$101A 91 02 STA ($02),Y


The comparison is OK, I think. If the contents of register A and the contents of the address $00 are equal then you load 1 into A and store the contents of A (1) into DEEK(2)+Y... therefore something like POKE (DEEK(2)+Y),A or, as Y=0; POKE DEEK(2),A
Here is important. You said "If the contents of register A and the contents of the address $00 are equal then you load 1 into A" I want to load content of (1) into A. So What I want to do is IF PEEK(DEEK(2)) = PEEK(0) THEN POKE DEEK(2),PEEK(1) . So, I think this part is also correct..

After changes according to your suggestions, my code is like that

Code: Select all

$1000  A9 00      LDA #$00      .. 
$1002  85 02      STA $02       .. 
$1004  A9 A0      LDA #$A0      .. 
$1006  85 03      STA $03       .. 
$1008  A9 00      LDA #$00      .. 
$100A  85 04      STA $04       .. 
$100C  A9 1F      LDA #$1F      .. 
$100E  85 05      STA $05       .. 
$1010  A0 00      LDY #$00      .. 
$1012  B1 02      LDA ($02),Y   .. 
$1014  C5 00      CMP $00       .. 
$1016  D0 04      BNE $101C     .. 
$1018  A5 01      LDA $01       .. 
$101A  91 02      STA ($02),Y   .. 
$101C  E6 02      INC $02       .. 
$101E  D0 02      BNE $1022     .. 
$1020  E6 03      INC $03       .. 
$1022  C6 04      DEC $04       .. 
$1024  A5 04      LDA $04       .. 
$1026  C9 FF      CMP #$FF      .. 
$1028  D0 02      BNE $102C     .. 
$102A  C6 05      DEC $05       .. 
$102C  A5 04      LDA $04       .. 
$102E  05 05      ORA $05       .. 
$1030  D0 E0      BNE $1012     .. 
$1032  60         RTS           
`

Aaaaaand it worked perfectly :)

But It is a bit slower than I expected. Do you have any suggestion to make it faster? Perhaps instead of counting down from 8000 to zero to check if we reach end of HIRES, we can check the DEEK(2). How?

Also, as you see, I am not using Y register and instead increase two byte the zero page address. Do you think it goes faster if we use Y register to scan each 256 byte and then increase 1 byte in (3)?

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Re: A little help for a simple machine code program.

Post by Dbug » Mon Jul 29, 2013 9:29 pm

If you are fine with overshooting a bit (ie: also do the replacement in the three lines of TEXT as well), you could indeed optimize by using both Y and a single 8 bit counter: 32 times 256=8192, so you could just set the address to $A000, and increment Y then increment the high byte each time Y reaches back to zero.
That would be wayyyyy faster :)

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Re: A little help for a simple machine code program.

Post by Chema » Mon Jul 29, 2013 9:38 pm

Yep, and if the counter is stored in register X, it would be even faster. Initialize it to 32, then decrement at each loop. A dex:bne $1012 would do.

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Re: A little help for a simple machine code program.

Post by peacer » Mon Jul 29, 2013 10:48 pm

Thanks :) I'll try to implement these advices to the code.

I am making a new "surprise" game using lightpen. These helps will make it better . I hope tomorrow it will be ready to play :D

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