Learning Machine Code

[PeterJ]

I've mentioned this previously, but this three part tutorial from [mention]Shaun_B[/mention] really gave me the introduction I needed. It originally appeared in MicroMart magazine.

http://shaunbebbington.blogspot.com/201 ... i.html?m=1

You just need Pasmo which is an opensource command line assembler, notepad and an emulator.

[Hedge1970]

I've mentioned this previously, but this three part tutorial from @Shaun_B really gave me the introduction I needed. It originally appeared in MicroMart magazine.

http://shaunbebbington.blogspot.com/201 ... i.html?m=1

You just need Pasmo which is an opensource command line assembler, notepad and an emulator.

Excellent and thank you I will also add this to the list of learning tools. Also thanks again to joefish, I had not realised zxspin could be run from command line so I have now edited the batch file run from Notepad++ to start Z Spin and load in the newly created .tap file from Pasmo. My Batch file is called RunPasmo.bat

Code: Select all

@ECHO OFF
REM NotePad RUN command "drive:\filepath\RunPasmo.bat $(FULL_CURRENT_PATH)"
REM $(FULL_CURRENT_PATH) is a command from Notepad++ for full path to current file

REM %~n1 Extracts file name without the type from $(FULL_CURRENT_PATH)
REM %dp1 Extracts just the file path from $(FULL_CURRENT_PATH)

REM Pasmo command line
E:\personal\computer\ZXSpectrum48K\programing\pasmo-0.5.3\pasmo-0.5.3\pasmo.exe --name %~n1 --tapbas %~dp1\%~n1.asm %~dp1\%~n1.tap 

REM Make new directory in the path wher the .asm file is called TAP
MKDIR %~dp1\TAP

REM Put the new created .tap file in the directory TAP
MOVE %~dp1\*.tap %~dp1\TAP

REM Launch Spin
E:\personal\computer\ZXSpectrum48K\emulator\zxspin_0666emu\ZXSpin.exe %~dp1Tap\%~n1.tap

[Hedge1970]

I have a question about knowing where in your code subroutines are. It really harks back to OpCodes and Assemblers vs Directly entering Hex. I expect I know the answer but I find it hard to believe

So if you look at page 93 of Mastering Machine Code -> https://spectrumcomputing.co.uk/index.p ... id=2000237 here Toni Baker writes an example routine for scrolling the screen up or down. I have published my .asm file below. Please note I added the ORG 32768.

Code: Select all

ORG 32768
UPLD	PUSH BC 
		PUSH DE
		PUSH HL
		LDIR 
		POP HL
		INC H
		POP DE
		INC D
		POP BC
		RET
UP  	LD HL,#4020
		LD DE,#4000
		LD BC,#1AE0
		LDIR
		LD HL,#47E0
		LD DE,#40E0
		LD C,#20
		LD A,#10
U_LOOP CALL UPLD
		DEC A
		JR NZ,U_LOOP
		LD H,D
		LD E,#E1
		DEC BC
		LD A,#08
U_BLANK LD (HL),#00
		CALL UPLD
		DEC A
		JR NZ,U_BLANK
		LD H,#5A
		LD D,H
		LD A,(#5C48);bordcr
		LD (HL),A
		LDIR
		RET 
DOWNLD 	PUSH BC
		PUSH DE
		PUSH HL
		LDDR
		POP HL
		DEC H
		POP DE
		DEC D
		POP BC
		RET
DOWN 	LD HL,#5ADF
		LD DE,#5AFF
		LD BC,#1AE0
		LDDR
		LD HL,#501F
		LD DE,#571F
		LD C,#20
		LD A,#10
D_LOOP 	CALL DOWNLD
		DEC A
		JR NZ,D_LOOP
		LD H,D
		LD E,#1E
		DEC BC 
		LD A,#08
D_BLANK LD (HL),#00
		CALL DOWNLD
		DEC A
		JR NZ,D_BLANK
		LD H,#58
		LD D,H
		LD A,(#5C8D);ATTR_P
		LD (HL),A
		LDDR
		RET  

She then provides the Basic program to use the Machine Code. In the Basic Program depending on keyboard key "7" or "6" being pressed - lines 160 and 150 - the Machine Code subroutines UP or DOWN are called. Of course she has not instructed you where (memory location) to build the Machine code to - in my case I used 32768 - which means she can only tell you where in the code sub-routine up and sub-routine down are by referencing the labels UP and DOWN so she states USR up and USR down at the relevant points. In order to get it working I added the reference USR 32841 and USR 32779 at the correspond position in Basic, but I only know these locations by actually counting the Bytes used in the code - referencing page 93 above I literally count the bytes from 32768 onwards until I reach Label UP and Label Down.

The question is - Is there a way of knowing by looking at the OPcode above what the byte count is to a specific point in your code or do you just need to keep track? If you need to keep track how would you do this in a large program like ManicMiner or Elite?

I do hope that makes sense.

Code: Select all

10 DIM A$(22,36) 
20 FOR I = 1 TO 22 
30 LET B$ = CHR$ (INT (96*RND) + 32) 
40 FOR J = 0 TO 5 
50 LET B$ = B$ + B$ 
60 NEXT J 
70 LET A$(I) = CHR$ 16 + CHR$ INT (8*RND) + CHR$ 17 + CHR$ INT(8*RND) + B$ 
80 PRINT A$(l) 
90 NEXT I 
100 LET A = 1 
110 PAUSE 0 
120 LET B$ = lNKEY$ 
130 LET B = A + 1 :IF B = 23 THEN LET B = 1 
140 LET C = A – 1 :IF C = 0 THEN LET C = 22 
150 IF B$ = "6" THEN PRINT AT 0,USR down;A$(C): LET A = C 
160 IF B$ = "7" THEN PRINT AT 21,USR up; A$(A): LET A = B 
170 INPUT ""
180 GO TO 110 

[Joefish]

PASMO has an option to let you save out the 'Symbol Table' during compilation.
This dumps the value of every variable (and thus location of every label), in alphabetical order, to a text file at the end of compilation.

I sometimes use it to calculate the size of the code or data by putting a label at the beginning and end, calculating a variable from the difference, and giving it a name like a taxi firm ("___A1A!!!AAA_TAXIS") so it appears high up in the alphabetical list.

(Which reminds me of the most obscure joke in an animation ever seen - the sign for "Thigmo Taxis" on the side of a building during a chase scene in Little Rodentia in Disney's Zootropolis / Zootopia. "Thigmotaxis" is the tendency of rodents to run alongside walls instead of out in the open. Rather than being some sort of rodent agoraphobia, they've actually evolved a reassuring tactile sensation ('thigmo' = 'tactile') when moving along a wall ('taxis' = 'movement', Ancient Greek) on one or the other side of their body; it's quite literally how they get from one place to another).

[Hedge1970]

Thanks Joefish, I’ll take a look later.

[AndyC]

Another trick is to start your code with a series of JP instructions to the entry points. Since each is three bytes long the addresses are easily predictable and won't move each time you reassemble the listing.

[Hedge1970]

That should work really well for this. Thanks

[Hedge1970]

Hi All,

So I have been busily working my way through some examples. Feeling quite good but with lots to learn and then today I got hit with something that I had assumed I knew …

ORG 32768

I thought that this statement meant that my code would start at the address 32768 and then in order to execute that code I then input RANDOMIZE USR 32768. So this all works, but tonight I needed to shift the code 54bytes higher so I did

ORG 32822 and then RANDOMIZE USR 32822 and I get an error - same code as before. Then when I type RANDOMIZE USR 32768 it all runs as expected??? Have I miss understood the ORG and if so how would I decide where to put machine code in the memory?

For reference here is the code. Not very exciting and needs a CLR before running...

Ideally it requires
700 RANDOMIZE USR 32768
710 INPUT LINE A$
720 RANDOMIZE USR 32781

Then it allows you to enter co-ords - nothing happens except you can enter more co-ords the actual move sub will come once I can relocate the code to a new part of the machine i.e. 32822

Thanks for any and all input.

Code: Select all

ORG 32768
			JP start		; 3 BYTES
			JP move			; 3 BYTES
s_print		POP HL			; pulls address off stack
			LD A,(HL) 		; This refers to first byte of defm
			INC HL			; alter +1 return byte address
			PUSH HL
			AND A			; Reset Carry and Set Zero if 0
			RET Z			; when we reach defb #00 return to JP board
			RST #10 		; prints message to screen
			JR s_print
start  		XOR A			; A=0, Reset Carry
			LD (#5C3C),A	; TVFlag
			CALL s_print	; pushes the address of next instruction i.e. the space below onto stack
			DEFM  " 12345678",#0D
			DEFM  "1",#90," ",#90," ",#90," ",#90," 1",#0D
			DEFM  "2 ",#90," ",#90," ",#90," ",#90,"2",#0D
			DEFM  "3        3",#0D
			DEFM  "4        4",#0D
			DEFM  "5        5",#0D
			DEFM  "6        6",#0D
			DEFM  "7",#90," ",#90," ",#90," ",#90," 7",#0D
			DEFM  "8 ",#90," ",#90," ",#90," ",#90,"8",#0D
			DEFM  " 12345678"
			DEFB #00		;sets the byte to 0 - end sub routine s_print
			JP board
row			LD A,#04
row_2		LD (HL),C		;color next draughts piece
			INC HL
			INC HL
			DEC A
			JR NZ,row_2
			ADD HL,DE		;next row
			BIT #0,L
			JR NZ,row_3
			INC HL
			INC HL
row_3		DJNZ row
			RET
board		LD HL,#5800		;PrintAt 0,0 attribute
			LD DE,#0016		
			LD C,#0A
yellow_1	LD B,#0A		;B=10 row 2 yellow, 4 black 4 white  
yellow_2	LD (HL),#30
			INC HL
			DJNZ yellow_2
			ADD HL,DE
			DEC C
			JR NZ,yellow_1
			LD HL,#5821
			LD E,#18
			LD C,#08
white_1		LD B,#08
white_2		LD (HL),#38
			INC HL
			DJNZ white_2
			ADD HL,DE
			DEC C
			JR NZ,white_1
			LD HL,#5822
			DEC E
			LD BC,#0307
			CALL row
			LD BC,#0200
			CALL row
			LD BC,#0302
			CALL row		
			RET
move		LD HL,(#5C4B)	;VARS system variable
			INC HL
			LD E,(HL)
			INC HL
			INC HL
			LD A,(HL)
			CP #E2		;STOP Sym A
			JR NZ,a$_ok
			RST #08
			DEFB #FF
a$_ok		LD HL,#02C6
			LD (#5C42),HL
			LD A,#01
			LD(#5C44),A
			RET

[Ast A. Moore]

Assembler directives and pseudo-instructions are pretty specific to a particular flavor of assembler, so it’s hard to tell without knowing your setup. I quickly tried your code in zasm and, after a tiny bit of cleanup, it worked fine. I could easily relocate your code anywhere in RAM (not that anything would suggest I couldn’t).

[Hedge1970]

Of course should have realised. I am using Pasmo as the assembler and and standard text editor to write the assembly. It’s running through ZX Spin.

Thanks for running the code and glad it worked I am now thinking I should use hex instead of decimal in the ORG statement - I’ve had issues in the code with decimal.

It’s all shut down now but I will try it tomorrow.

[catmeows]

Why USR 32781 ? Where it should jump ?
Anyway, I'm not seeing any issues with your code. Still, I would not use $ in label, it can have special meaning.

[Ast A. Moore]

I am using Pasmo as the assembler and and standard text editor to write the assembly. It’s running through ZX Spin.

I see. Well, I’m not familiar with either, but off the top of my head, I’d suggest putting a space in front of the ORG statement. Some assemblers assume that a letter at the first position of a line is a label, rather than a (pseudo-) instruction/directive.

[lister_of_smeg]

I see. Well, I’m not familiar with either, but off the top of my head, I’d suggest putting a space in front of the ORG statement. Some assemblers assume that a letter at the first position of a line is a label, rather than a (pseudo-) instruction/directive.

Yup, this. And as this results in you having no defined origin, I would hazard a guess that Pasmo defaults to 32768 in such a case.

[Hedge1970]

Hey I did not see these last night and tried the hex before coming to work but that failed.

Will give this a shot tonight

[uglifruit]

Just as a test it appears to work okay for me. Should the you mention 32781 be 32771 ? (to the second jp command) - though this gives a N Statement Lost Error.

Relocating the routine to a different address (different org instead of 32768), and calling it there seems to work the same.

I'm using PASMO on a mac: with these arguments:
pasmo --tapbas try/try.asm try/try.tap

Just as a 'did you know' - finishing your assembly with END 32768 will compile the basic loader so it already does a Randomize 32768 when you launch it in your emulator.

[Alessandro]

Yup, this. And as this results in you having no defined origin, I would hazard a guess that Pasmo defaults to 32768 in such a case.

It doesn't. If you do not specify any code beginning address with ORG in the source, Pasmo will produce, when assembling to a tape image file, a Bytes header stating address 0 as the first location, and all relative jumps in the code will take that into account.

[Hedge1970]

I see. Well, I’m not familiar with either, but off the top of my head, I’d suggest putting a space in front of the ORG statement. Some assemblers assume that a letter at the first position of a line is a label, rather than a (pseudo-) instruction/directive.

....Just as a 'did you know' - finishing your assembly with END 32768 will compile the basic loader so it already does a Randomize 32768 when you launch it in your emulator.

Thank you Gents,

Not only did the space before ORG do the trick, but I got an awsome tip from Uglifruit as well.

Back to trying to understand whats going on

[Hedge1970]

I remember years ago as a teenager people said you needed to be good at maths to be a programmer. Many years later I quite happily did web work, SQL, C#, DotNet etc and never found that maths played a significant part.

Now I get it, hex to dec to bin plus the relationships to the screen and attributes to hex and bin rra rrl wtf;). I’ve not yet got settled with understanding these relationships but I can see them and I can work my way through them but they don’t come naturally.

[Ast A. Moore]

Now I get it, hex to dec to bin plus the relationships to the screen and attributes to hex and bin rra rrl wtf;). I’ve not yet got settled with understanding these relationships but I can see them and I can work my way through them but they don’t come naturally.

That’s probably because there’s nothing inherently natural about hexadecimal or binary. Come to think of it, this is also true for the decimal system; we’re just familiar with it. (Well, we do have ten fingers and toes, but then there are cultures that don’t use base 10 counting systems.)

It’s totally fine to mix and match different counting systems throughout your code. I do it all the time. Binary is useful when working with bits (shifts, masks, etc.), hexadecimal is more suited for calculating addresses or working with 16–bit-wide registers, decimal is convenient for working out simple coordinates, score, and other “regular” values.

Advanced math is probably non-essential, but some basic math skills always come in handy.

Here’s a small real-life example of code that uses all three bases:

Code: Select all

	ld a,%01000111	;ATTR: no FLASH, BRIGHT, BLACK PAPER, WHITE INK
	ld hl,$5907	;ATTR address
	ld (hl),a
	inc l
	ld (hl),a

	ld b,6		;counter for a short pause
2$	halt
	djnz 2$

[Hedge1970]

Hi,

Its been quite a journey over the last few weeks and while there is no light on the horizon just yet, I can at least understand around 70% of the work I am copying. I am making my way through Mastering Machine Code (https://ia800604.us.archive.org/view_ar ... ectrum.pdf)

I am using Spin emulator and Pasmo compiler, whereas in the book the author Toni Baker provides her own machine code writing program and herein lies one of the challenges, that is trying to bridge the gap between what she says to do and what to do in my Pasmo/Spin environment.

Most of the examples and programs have been ok and with a lot of thought and a little tweak I manage to get them running, however this little game has me foxed. Its Chapter 15, Page 243 via the above link.

The code below is very similar to the book and while it may not be 100% the same, the issue I am asking about is perfectly working and copied below - the layout of the "spiral". It looks nothing like a spiral and - I hope I am correct here - the loop to work through the parts of the SPIRAL DEFW statements ( LD B,#0A) is only looping 10 times (0A) whereas there are 20 bytes of data so the program ignores the last 10 bytes? This code is three lines below the START label. For the DEFW statements Ive used both binary and hex and both behave the same way.

Code: Select all

			ORG 26621
			JP start
spiral		DEFW #FF,#E0
			DEFW #A0,#20
			DEFW #AF,#A0
			DEFW #A8,#A0
			DEFW #AA,#A0
			DEFW #AE,#A0
			DEFW #A0,#A0
			DEFW #BF,#A0
			DEFW #80,#20
			DEFW #FF,#E0
score_1		DEFM #0D
			DEFM "Your "
			DEFM "Score "
			DEFM "Now "
			DEFM "999000"
score_c		DEFM #16,#0B,#0F
			DEFM "000"
start		XOR A
			LD (#5C3C),A
			LD B,#0A
			LD HL,#6800
s_loop		LD E,(HL)
			INC HL
			LD D,(HL)
			INC HL
			EX DE,HL
r_loop		ADD HL,HL
			JR C,wall
			LD A,#3F
			LD (#5C8F),A		;used A to ld 3F
			JR prsq
wall		LD A,#00
			LD (#5C8F),A		;used A to ld 00
prsq		LD A,#2B
			RST #10
			LD A,H
			OR L
			JR NZ,r_loop
			LD A,#0D
			RST #10
			EX DE,HL
			DJNZ s_loop
			LD A,#38
			LD (#5821),A
			LD A,#30
			LD (#5C8F),A	;Used A to ld 30
			LD HL,#6828		;score_1
			LD B,#15
sc_loop		LD A,(HL)
			INC HL
			RST #10
			DJNZ sc_loop
			LD HL,#3939
			LD(#6841),HL	;score c+3
			LD(#6842),HL	;score c+4
			LD HL,#5821
			LD(#5C92),HL
			LD HL,#0000
			LD(#5C94),HL
loop		LD HL,#6843		;score_c + 5
deci		LD A,(HL)
			CP #0F
			JR NZ,positive
			LD B,#03
reset		INC HL
			LD (HL),#30
			DJNZ reset
			RET
positive	DEC A
			CP #2F
			JR NZ,ok
			LD (HL),#39	;REPLACE 
			DEC HL
			JR deci
ok			LD (HL),A
			LD HL,#683E
			LD B,#06
cs_loop		LD A,(HL)
			INC HL
			RST #10
			DJNZ cs_loop
			LD BC,#0800
delay		DEC BC
			LD A,B
			OR C
			JR NZ,delay
			CALL #028E
			LD A,E
			CP #09
			JR Z,left
			CP #10
			JR Z,down
			CP #11
			JR Z,right
			CP #12
			JR NZ,loop
up			LD DE,#FFE0
			JR move
left		LD DE,#FFFF
			JR move
down		LD DE,#0020
			JR move
right		LD DE,#0001
move		LD HL,(#5C94)
			LD A,H
			OR L
			JR Z,move_ok
			ADD HL,DE
			LD A,H
			OR L
			JR NZ,loop
move_ok		LD HL,(#5C92)
			LD A,(HL)
			XOR #07
			LD (HL),A
			ADD HL,DE
			LD A,(HL)
			XOR #07
			LD (HL),A
			LD (#5C92),HL
			LD HL,#0000
			CP #38
			JR Z,finish
			LD H,D
			LD L,E
finish		LD (#5C94),HL
			LD HL,(#5C92)
			LD DE,#5885
			SBC HL,DE
			RET Z
			JP loop

In my many many experiments with the code I was able to make this slightly modified version. However I thought I would make use of all 20 bytes and added a new loop called SET_C. But when I implement it I get two spirals but not joined side by side, rather separated. I have looked as hard as I can to understand why and realise that I have no understanding of how the first position of the spiral is .. positioned. If anyone can find it in themselves to help I would love to know, then I will/should be able to create a bigger spiral game screen.

Note in both above and below code I have replaced her #5536 for #5C8F to target the Speccy System Variable (ATTR T)

My Code

Code: Select all

			ORG 26621
			JP start
score_1		DEFM #0D,"Your "
			DEFM "score "
			DEFM "now " 
			DEFM "999000"				;22 letters and spaces
score_c		DEFM #16,#0B,#0F			;3
			DEFM "000"					;3
spiral		DEFW %11111111
			DEFW %00000001
			DEFW %11111101
			DEFW %11000101
			DEFW %10010101
			DEFW %10110101
			DEFW %10100101
			DEFW %10111101
			DEFW %10000001
			DEFW %11111111				;10 start 26644
start		XOR A
			LD (#5C3C),A
			LD B,#0A
			LD HL,#681C					;address of spiral data
set_c		LD C,#01
s_loop		LD E,(HL)
			INC HL
			LD D,(HL)
			INC HL
			EX DE,HL
r_loop		ADD HL,HL
			JR C,wall
			LD A,#3F
			LD (#5C8F),A				;ATTR_T 0011 1111 FL BR P2 P1 P0 I2 I1 I0
			JR prsq
wall		LD A,#00
			LD (#5C8F),A				;ATTR_T 0000 0000 FL BR P2 P1 P0 I2 I1 I0
prsq		LD A,#2B					;+ character
			RST #10
			LD A,H
			OR L
			JR NZ,r_loop
			EX DE,HL
			LD A,C
			DEC C
			CP #01
			JR NZ,s_loop
			LD A,#0D						;return new line
			RST #10
			DJNZ set_c
			LD A,#38
			LD (#5821),A
			LD A,#30
			LD (#5C8F),A					;Black on yellow
			LD HL,#6800						;score_1
			LD B,#15
sc_loop		LD A,(HL)
			INC HL
			RST #10
			DJNZ sc_loop
			LD HL,#3939
			LD (#6819),HL					;score_c +3
			LD (#681A),HL					;score_c +4
			LD HL,#5821
			LD(#5C92),HL					;Store current position
			LD HL,#0000
			LD(#5C94),HL 					;Store last move
loop		LD HL,#681B						;Score_c + 5
decemal		LD A,(HL)
			CP #0F
			JR NZ,positive
			LD B,#03
reset		INC HL
			LD (HL),#30 					;"0"
			DJNZ reset
			RET
positive	DEC A
			CP #2F
			JR NZ,ok
			LD (HL),#39						;"9" 
			DEC HL
			JR decemal
ok			LD (HL),A
			LD HL,#6816 					;score_c
			LD B,#06
cs_loop		LD A,(HL)
			INC HL
			RST #10
			DJNZ cs_loop
			LD BC,#0F00						;speed
delay		DEC BC
			LD A,B
			OR C
			JR NZ,delay
			CALL #028E						;Call KEYSCAN
			LD A,E
			CP #09
			JR Z,left
			CP #10
			JR Z,down
			CP #11
			JR Z,right
			CP #12
			JR NZ,loop
up			LD DE,#FFE0
			JR move
left		LD DE,#FFFF
			JR move
down		LD DE,#0020
			JR move
right		LD DE,#0001
move		LD HL,(#5C94)					;is player in wall?
			LD A,H
			OR L
			JR Z,move_ok
			ADD HL,DE
			LD A,H
			OR L
			JR NZ,loop
move_ok		LD HL,(#5C92)
			LD A,(HL)
			XOR #07
			LD (HL),A
			ADD HL,DE
			LD A,(HL)
			XOR #07
			LD (HL),A
			LD (#5C92),HL
			LD HL,#0000
			CP #38
			JR Z,finish
			LD H,D
			LD L,E
finish		LD (#5C94),HL				;Last move
			LD HL,(#5C92)
			LD DE,#58CB					;Square at center
			LD A,#A7
			LD (DE),A					;10100111 FL BR P2 P1 P0 I2 I1 I0
			SBC HL,DE
			RET Z
			JP loop

[Ast A. Moore]

DEFW is an assembler pseudoinstruciton that defines a word, that is a sixteen–bit-wide (i.e. two-byte) space. Different assemblers will treat its argument differently, if it’s not specified fully. In your case, DEFW #FF,#E0 actually means “write the following four bytes: $ff $00 $ef $00,” provided the assembler just fills out the unspecified byte with zeros.

What you probably want is to rewrite all those statements to read something like DEFW #FFE0, without the commas.

However, note that DEFW #FFE0 will probably store the data LSB-first, i.e. #E0 #FF, and not #FF #E0.

DEFB defines a byte, and DEFM is used by some assemblers to define a message (or string).

Most assemblers allow you use a comma to separate the arguments to save space:

DEFW #fd30,#e400,#21a0
DEFB 254,1,33,14
DEFM “string”, $20, “another string”

A lot of the times, the DEFB and DEFM statements are completely interchangeable.

There’s also DEFS, which defines (reserves) a space of a specific length and can (optionally) fill it with a specific value:

DEFS 30 $FE means, fill the next thirty bytes with the value $FE.

[Hedge1970]

DEFW is an assembler pseudoinstruciton that defines a word, that is a sixteen–bit-wide (i.e. two-byte) space. Different assemblers will treat its argument differently, if it’s not specified fully. In your case, DEFW #FF,#E0 actually means “write the following four bytes: $ff $00 $ef $00,” provided the assembler just fills out the unspecified byte with zeros.

Oh my goodness I thought I had tried every combination... This works And it places the spiral at 0,0 which is where I was expecting it to be

Thank you so much.

[Ast A. Moore]

Oh, a small optimization technique. Whenever possible, try avoiding the SBC HL,DE instruction. It takes 15 T states to execute and, unless you want it, requires that you clear the carry bit. What it does is it subtracts both DE and Carry from HL. If Carry is already zero, then it’s just a regular subtraction; if Carry is set, then an extra one will be subtracted from HL. So, if you don’t want that, you’ll have to precede it with an OR A—another 4 T states and an extra byte.

There’s no “subtract a 16-bit value without carry” instruction, unfortunately. However, there’s a workaround. If the value you want to subtract from HL is a constant, simply change its sign from “+” to “–” and use ADD HL,DE (only 11 T states).

So, instead of:

Code: Select all

	ld de,2343
	or a
	sbc hl,de

write this:

Code: Select all

	ld de,-2343
	add hl,de

That, of course, depends on whether the assembler can understand this notation. Still, even if it can’t you can convert your number yourself by subtracting it from $10000.

[Hedge1970]

I have seen certain discussions on optimisation and T-states, for now I am focused on getting a broad understanding of the key terms and how to structure a program. Its quite tricky when yoiur limited to the few registers to create looping structures. In higher level languages I would just make/add variables as and when needed, but here its a careful balance.

It may seem like a small thing to you but I am so grateful for the reply as I dont have many friends learning 8bit machine code

Here is the finished product albeit still with the now not needed SET_C loop I added.

Code: Select all

			ORG 26621
			JP start
score_1		DEFM #0D,"Your "
			DEFM "score "
			DEFM "now " 
			DEFM "999000"				;22 letters and spaces
score_c		DEFM #16,#0B,#0F			;3
			DEFM "000"					;3
spiral		DEFW #FFE0
			DEFW #A020
			DEFW #AFA0
			DEFW #A8A0
			DEFW #AAA0
			DEFW #AEA0
			DEFW #A0A0
			DEFW #BFA0
			DEFW #8020
			DEFW #FFE0;10
start		XOR A
			LD (#5C3C),A
			LD B,#0A
			LD HL,#681C					;address of spiral data
set_c		LD C,#01
s_loop		LD E,(HL)
			INC HL
			LD D,(HL)
			INC HL
			EX DE,HL
r_loop		ADD HL,HL
			JR C,wall
			LD A,#3F
			LD (#5C8F),A				;ATTR_T 0011 1111 FL BR P2 P1 P0 I2 I1 I0
			JR prsq
wall		LD A,#00
			LD (#5C8F),A				;ATTR_T 0000 0000 FL BR P2 P1 P0 I2 I1 I0
prsq		LD A,#2B					;+ character
			RST #10
			LD A,H
			OR L
			JR NZ,r_loop
			EX DE,HL
			LD A,C
			DEC C
			CP #01
			JR NZ,s_loop
			LD A,#0D						;return new line
			RST #10
			DJNZ set_c
			LD A,#38
			LD (#5821),A
			LD A,#30
			LD (#5C8F),A					;Black on yellow
			LD HL,#6800						;score_1
			LD B,#15
sc_loop		LD A,(HL)
			INC HL
			RST #10
			DJNZ sc_loop
			LD HL,#3939
			LD (#6819),HL					;score_c +3
			LD (#681A),HL					;score_c +4
			LD HL,#5821
			LD(#5C92),HL					;Store current position
			LD HL,#0000
			LD(#5C94),HL 					;Store last move
loop		LD HL,#681B						;Score_c + 5
decemal		LD A,(HL)
			CP #0F
			JR NZ,positive
			LD B,#03
reset		INC HL
			LD (HL),#30 					;"0"
			DJNZ reset
			RET
positive	DEC A
			CP #2F
			JR NZ,ok
			LD (HL),#39						;"9" 
			DEC HL
			JR decemal
ok			LD (HL),A
			LD HL,#6816 					;score_c
			LD B,#06
cs_loop		LD A,(HL)
			INC HL
			RST #10
			DJNZ cs_loop
			LD BC,#0F00						;speed
delay		DEC BC
			LD A,B
			OR C
			JR NZ,delay
			CALL #028E						;Call KEYSCAN
			LD A,E
			CP #09
			JR Z,left
			CP #10
			JR Z,down
			CP #11
			JR Z,right
			CP #12
			JR NZ,loop
up			LD DE,#FFE0
			JR move
left		LD DE,#FFFF
			JR move
down		LD DE,#0020
			JR move
right		LD DE,#0001
move		LD HL,(#5C94)					;is player in wall?
			LD A,H
			OR L
			JR Z,move_ok
			ADD HL,DE
			LD A,H
			OR L
			JR NZ,loop
move_ok		LD HL,(#5C92)
			LD A,(HL)
			XOR #07
			LD (HL),A
			ADD HL,DE
			LD A,(HL)
			XOR #07
			LD (HL),A
			LD (#5C92),HL
			LD HL,#0000
			CP #38
			JR Z,finish
			LD H,D
			LD L,E
finish		LD (#5C94),HL				;Last move
			LD HL,(#5C92)
			LD DE,#5885				;Square at center
			LD A,#A7
			LD (DE),A					;10100111 FL BR P2 P1 P0 I2 I1 I0
			SBC HL,DE
			RET Z
			JP loop

[Ast A. Moore]

You need to use labels properly, otherwise you’ll get confused by your own code in no time flat.

For example:

Code: Select all

LD HL,#681C		;address of spiral data

What if you want to move that data (and you will, if some other data comes before it)? Since you’ve already defined it (by assigning a label to it), just tell the assembler that:

Code: Select all

LD HL,spiral

The same goes for the rest of the data:

Code: Select all

LD (#6819),HL		;score_c +3

can be replaced with

Code: Select all

LD (score_c+3),HL

and so on.

That way, you can move your actual data somewhere else in your code, say, append it to the end, and avoid the unnecessary JP start instruction. In addition, your ORG statement doesn’t have to be meticulously calculated each time. Just stick an easy to type number there, like 40000.

I haven’t looked at your code with any diligence, but the penultimate instruction is RET Z, right before a JP. I don’t see any CALLs in your code (aside from a single ROM CALL). Where exactly is you code supposed to return to once it gets there and the condition is true?