Home > Z80 Assembly > Sprite Graphics Tutorial

Sprite Graphics Tutorial

This tutorial is by Derek M. Smith © 2005.

Sprite graphics are the backbone of arcade games. Put simply, a sprite is a moveable screen object, such as a spaceship, alien, or anything else you can imagine. Some computers come with built in sprites, such as the Commodore and Atari ranges. These machines could generate sprites without the need for complex programming, and on top of this, they used an additional processor, which left the main CPU free to get on with the rest of the program. The Spectrum sadly was never adorned with such luxuries and so programmers had to write their own sprite routines from scratch (a daunting task for the amateur).

Sloppy programming in this area could really spoil a game, either making it too slow, or through flickering graphics.

A simple sprite routine is often one of the first things a new programmer, after having mastered the basics of machine code, will have a go at.

At first it seems straight forward, but soon all sorts of hurdles appear. The first usually being the Spectrum’s unusual screen arrangement. More experienced programmers who have mastered the basic sprite handling routine will then seek ways to optimise it.

This tutorial is aimed at both beginners and the more experienced programmers. Beginners will learn the principles behind sprite programming and the experienced machine coders will learn to find ways of improving their routines. For simplicity, this tutorial assumes that the reader has a decent grasp of assembly language.

ZX Spectrum screen arrangement
“The display file stores the television picture. It is rather curiously laid out…” -Spectrum Manual ch24 p164

On the face of it the Spectrum’s screen arrangement is pretty strange. Just watch the screen of a game loading, and you’ll understand what I mean. Why does it skip lines like that? Sure it may be interesting to watch the picture being gradually built up, but it can be a real pain when it comes to writing a sprite routine (at least until you understand the principles behind its layout).

First off, the Spectrum screen has a resolution of 256 pixels across by 192 down, not including the border around it. It can display 8 colours (including black and white) with two levels of intensity (brightness).

The Memory Map for the screen starts at address 16384 and is 6912 bytes long.
It is split in to two halves with the first 6144 bytes containing the bit-map (or pixel map, if you wish) and the remaining 768 bytes containing the attribute-map.

Consider the bit-map first:-

Each line of 256 pixels is stored as 32 bytes: 32 x 8 = 256. So far so good.

Now you would think that each line would follow on from the one before in the pixel map, and most people (myself included) write their first sprite routine thinking this to be the case – only to find that when they execute the routine, with a shudder of anticipation, their sprite is spread all over the screen. At this point some give up (or decide to write adventure games instead) baffled by the Spectrum’s idiosyncrasies.

Let me say now that when you do grasp the screen layout and the techniques used for addressing it, I think you will be glad that the designers done it that way.

Type in the following and run it:

	10 FOR S = 16384 TO 22527
	20 POKE S, 255
	30 NEXT S

This short BASIC program fills the screen. Although it is moving through memory sequentially, POKING 255 into each memory location, the screen fills up in a rather more esoteric manner. Run the program a few times and watch the pattern it traces. You’ll notice two things: One is the way it skips lines, another is that the screen is divided into three parts.

I said earlier that each line is stored as 32 bytes, with the first line beginning at 16384. Where does the second line start? Most of us assume it would start at 16384+32. In fact it starts at 16384+256 and 16384+32 takes us down to line 8. Remember that the screen is used to display text (characters) as well as graphics. So adding 32 actually takes us down one character row (8 screen lines).

In machine code 16bit addresses, such as 16384, are stored in high byte / low byte format. The high byte is equal to ADDRESS / 256 and the low byte is the remainder from the division, so 16384 in high/low byte format would be 64, 0 (16384 / 256 = 64 with no remainder). This means that when we increase the high byte of a 16 bit address by one, it is equivalent to adding 256. So when it comes to screen addressing all we need to do to move down a line is increase the high byte of the address. This can be done quite easily and quickly in machine code:

	ld hl,16384	;load the hl register pair with the address of the start of the display file
	inc h		;increment the high byte (4 tstates)

This is in fact much quicker than adding 32 where we would have to do the following:

	ld hl,16384
	ld a,l		;load the accumulator with the l register as we cannot add 
			;directly to the l register
	add a,32	;add 32
	ld l,a		;load the result back into the l register (total time 15 tstates)

The above works within a group of 8 lines, ie. 1 char line.

Now, as mentioned before, the other peculiar thing about the screen layout is that it is divided in to three parts – top, middle and bottom. Each third of the screen has 64 lines (or 8 character rows) and takes up 2048 bytes of memory. All that has been said so far applies only so long as we don’t cross from one third into another.

The whole matter becomes at good deal clearer if we look at the screen address in binary.

           High Byte                |               Low Byte

0   1   0   T   T   L   L   L          Cr Cr Cr Cc Cc Cc Cc Cc

I have used some abbreviations to make things a bit clearer:

T – these two bits refer to which third of the screen is being addressed: 00 – Top, 01 – Middle, 10 – Bottom

L – these three bits indicate which line is being addressed: from 0 – 7, or 000 – 111 in binary

Cr – these three bits indicate which character row is being addressed: from 0 – 7

Cc – these five bits refer to which character column is being addressed: from 0 – 31

The top three bits ( 010 ) of the high byte don’t change.

Calculating the screen address

The first task in putting a sprite on the screen is to translate the X,Y coords into a screen address.
There are two ways of doing this. One is to set up a look-up table which contains 192 addresses corresponding to each screen line. The other way is a bit more interesting and involves distilling the appropriate line, column and row bits that make up the address from the X and Y coords. Lets examine this way first.

Taking the Y coord first: This will be in the range 0 – 191, with 0 corresponding to the top of the screen. The lowest three bits indicate which line (within a character row) 0 – 7 we are dealing with. This is the same as the high byte of the screen address. The top two bits refer to which third of the screen we are dealing with:

			   Y Coord

		T   T   -   -   -   L   L   L

Let us assume that the B & C registers contain our X & Y coords. First we need to isolate the lowest three bits of the Y coord (C reg.), as follows:

	ld a,c
	and %00000111		;% indicates that the number following it is in binary format

We will use the HL register pair for the screen address. So next we transfer these three bits to the high byte (H reg.)

	ld h,a

In order to get the top two bits into the correct position we must shift them right three times, as follows:

	ld a,c
	rra			;rotate right accumulator
	rra
	rra

This shifts all the bits to the right three times, with the highest bit being reset after each shift. All that’s left to do at this stage is to clear the bits we don’t need, as follows:

	and %00011000

This must now be ORd with the H register.

	or h

Then the top three bits, which remain constant, are set as appropriate:

	or %01000000
	ld h,a			;load the result back into the H reg.

So now we have the high byte of the screen address.

The low byte of the address is composed of the following bits of both the X and Y coords:

   	  	 X Coord              |               Y Coord

  Cc Cc Cc Cc Cc   -   -   -         -   -   Cr Cr Cr   -   -   -

		Low Byte

  Cr Cr Cr Cc Cc Cc Cc Cc

The character column bits need to be shifted to the right three times.

	ld a,b			;B reg. holds the X coord
	rra
	rra
	rra
	and %00011111		; ensure the top 3 bits are clear
	ld l,a

Then the character row bits must be shifted left twice so that they correspond to the highest three bits of the low byte of the address (see earlier).

	ld a,c			;C reg. holds the Y coord
	rla			;rotate left accumulator
	rla
	and %11100000		;isolate the character row bits
	or l			;OR the result with the low byte
	ld l,a			;and place in L register

Right, that’s the difficult part done. We now have the screen address in HL. All that is left is to get the pixel position (0 – 7 from left to right) from the X coord (lowest three bits).

	ld a,b
	and %00000111

So the complete routine is as follows:

On Entry: B reg = X coord,  C reg = Y coord
On Exit: HL = screen address, A = pixel position

; Calculate the high byte of the screen addressand store in H reg.

	ld a,c
	and %00000111
	ld h,a
	ld a,c
	rra
	rra
	rra
	and %00011000
	or h
	or %01000000
	ld h,a

; Calculate the low byte of the screen address and store in L reg.

	ld a,b
	rra
	rra
	rra
	and %00011111
	ld l,a
	ld a,c
	rla
	rla
	and %11100000
	or l
	ld l,a

; Calculate pixel postion and store in A reg.

	ld a,b
	and %00000111

Using a Look-up Table

The other method of calculating the screen address is to use a table of pre-calculated addresses, and then use the Y coord to pick out the right one. This is a good deal quicker than the above method, albeit at the cost of 384 bytes of memory for the table. The look-up table contains the address of each line in the display file, 192 lines.

Routine to generate a Screen Address Table:

scradtab	equ 64000

gentab		ld de,16384
		ld hl,scradtab
		ld b,192

lineloop	ld (hl),e
		inc l
		ld (hl),d
		inc hl

		inc d
		ld a,d
		and 7
		jr nz,nextline
		ld a,e
		add a,32
		ld e,a
		jr c,nextline
		ld a,d
		sub 8
		ld d,a

nextline	djnz lineloop

		ret

Assuming again that on entry the B and C registers contain our X & Y coords.

	ld de,scradtab		;address of look-up table
	ld l,c			;Y coord in L register
	ld h,0
	add hl,hl		;multiply by two, as each address is 2 bytes
	add hl,de		;add to the start address of the table. HL is now at the
				;appropriate point in the table
	ld a,(hl)		;get low byte of screen address; store in A temporarily
				;so as not to corrupt HL
	inc l
	ld h,(hl)		;get high byte
	ld l,a			;HL now contains the address of the start of the line

	ld a,b			;calculate character column 0 - 31 from X coord
	and %11111000		;isolate appropriate bits
	rrca			;rotate right circular accumulator (faster than srl a)
	rrca
	rrca			;shifting three times is the same as dividing by eight
	add a,l			;add to low byte of address
	ld l,a			;put result back in L reg.

On exit HL will contain the screen address.

An even quicker way is to arrange the lookup table so that it is aligned with a 256 byte page boundary. This method separates the low byte and the high byte of the screen addresses, so in effect there are two tables one containing the low bytes of the address of each screen line and one containing the high bytes. Each must be aligned with a page boundary, ie. the start address of the table must be cleanly divisible by 256.

The above routine for generating the screen address table needs the following changes made to it:

lineloop	ld (hl),e
		inc h
		ld (hl),d
		dec h
		inc l

The table would be filled as follows:

	64000 - 64191		0....... 32....... 64....... 96....... 128.......		Low bytes of screen addresses

	64256 - 64447		64,65,66....71, 64,65,66....71, 64,65,66....71....	High bytes of screen addresses

Finding the start address of a particular line is then done as follows:

	ld h,scradtab/256	; high byte of start of table
	ld l,c			; C reg. contains the Y coord
	ld a,(hl)		; already we have found the low byte, so store in A temporarily
	inc h			; increasing high byte moves forward 256 bytes, and to the
	ld h,(hl)		; corresponding high byte of the screen address		
	ld l,a			; HL now contains the screen address for the start of the line

All that is then needed is to add on the column position, which is calculated in the same way as the previous routine.

Basic considerations

The main factors which need consideration when writing a sprite routine are:

1. Whether or not to use masked sprites
2. Whether or not to use pre-shifted sprites
3. What sizes of sprite to use and whether to use a different routine tailored for each size or a generic
routine which will produce various sizes of sprite

Writing a generic sprite routine which can handle many sizes and types of sprite, is generally shunned by most programmers, because the complexity of the routine has a negative impact on performance. It is perferable to write several routines tailored to specific sizes of sprite.

As per usual there is a trade-off between memory and speed, with the fastest routines using the most memory.

Masked sprites on average will take about 30% more time to process.

Pre-shifted sprites, which are sprites stored in 2, 4 or 8 different pixel positions, are by far the fastest type, but consume lots of memory.

An alternative to using pre-shifted sprites favored by some programmers, is to have a look-up table containing each bit pattern from 0 – 255 shifted into eight positions. In practice this takes up 4K of memory but this is usually less than is required when pre-shifting a large number of sprites. More about this later.

A Worked Example

It’s now time to take a look at how to code a basic sprite routine. The example below is for an 8 x 8 non-masked sprite. The sprite data (a small arrow pointer) is located at USR “A” – 65368. The routine shifts the data in real time. The sprite is XORed with the contents of the screen. You could also OR the sprite with the screen contents (see below for explanation of OR, XOR etc.) Real time rotation of sprite data is easiest acheived with small sprites (16 pixels or less wide), as the data can be stored in registers while being rotated.

The steps in the algorithm to draw the sprite could be stated as follows:

1. Retrieve X and Y coords.
2. Calculate Screen Address based on X,Y coords, using a look-up table
3. Calculate Bit Position from X coord (X coord AND 7 = bit position)
4. Retrieve a line of the sprite graphics data
5. Check if Bit Position is zero
6. If so there is no need to shift the sprite data, so skip the code which shifts the data (jump to step 8 )
7. Shift the sprite data according to its bit position
8. Put the line of sprite data on the screen
9. Adjust screen address for next line
10. Perform steps 4-9 until all lines have been drawn

		ORG 50000

SPRITE	DI			;Disable Interupts. Not strictly necessary in this
					;example as we are not redirecting the Stack.

		LD BC,(XPOS)		; First off, get the X & Y coords
					; and place them in B & C registers

		LD H,SCRADTAB/256	; this next section calculates the
		LD L,B			; screen address using a lookup table
		LD A,(HL)		; as explained earlier in the tutorial
		INC H
		LD H,(HL)
		LD L,A
		LD A,C
		AND 248
		RRCA
		RRCA
		RRCA
		ADD A,L
		LD L,A
		LD (SCRADD),HL	; store screen address for later
					; as HL is needed again

		LD A,C			; find which pixel position the sprite
		AND 7			; will be at and store it. We need this to
		LD (BITPOS),A		; know how many times to shift the sprite data

		LD HL,SPRGFX	; start address for the sprite graphic

		LD C,8			; The sprite is 8 lines tall

LINELOOP	LD E,(HL)		; load the E reg with the first line of
		INC L			; sprite data, and move forward

		PUSH HL		; preserve this address for later

		LD A,(BITPOS)		; Retrieve the pixel position
		OR A			; Quick way of testing if A is zero - Note: A is unaffected
		JR Z,SKIPROTATE	; If zero then no shifting is needed

		LD B,A			; loop counter for number of times to
					; rotate (shift) sprite data
		XOR A			; Quick way setting A to zero and
					; clearing the Carry Flag. The Carry Flag must be reset
					; as the rotate instructions will shift its contents
					; into the sprite data.

ROTATELOOP	RR E			; An extra register is needed for the
		RRA			; shifted sprite data. After the RR E is executed
					; rightmost bit of sprite data is shifted out to the Carry Flag.
					; It is then shifted into the A reg by RRA. So no data is lost.
		DJNZ ROTATELOOP	; Loop back until shifting is complete

SKIPROTATE	LD D,A			; Store in D register, as A will be
					; needed for another purpose. E & D regs now contain the
					; shifted sprite data.

		LD HL,(SCRADD)	; Get back the screen address

		LD A,(HL)		; Actually put a line of the sprite on the screen
		XOR E			
		LD (HL),A
		INC L
		LD A,(HL)
		XOR D
		LD (HL),A

		DEC L			; move back and down one line
		INC H

		LD A,H			; It is not necessary to recalculate
		AND 7			; the screen address for each line of
		JR NZ,A1		; the sprite. All that is needed is to
		LD A,L			; check if a char. or segment boundary
		ADD A,32		; has been crossed and adjust address
		LD L,A			; accordingly.
		JR C,A1
		LD A,H
		SUB 8
		LD H,A

A1		LD (SCRADD),HL	; store it again

		POP HL		; retrieve the address of the next line
					; of sprite data
		DEC C		
		JP NZ,LINELOOP	; loop back until all lines are drawn.

		EI			; Enable Interupts again
		RET

XPOS		DEFB 0
YPOS		DEFB 0
SCRADD	DEFW 0
BITPOS	DEFB 0

SCRADTAB	EQU 64000

On Exit : A, BC, DE, HL corrupt

As can be seen from the above routine it is not necessary to re-calculate the screen address for every line of a sprite. A couple of checks can be made to test if the next line to be drawn is in a new character row or screen third and adjust the address accordingly:

	LD A,H				;HL contains screen address
	AND 7
	JR NZ, REST_OF_PROGRAM
	LD A,L
	ADD A,32
	LD L,A
	JR C, REST_OF PROGRAM
	LD A,H
	SUB 8
	LD H,A

Rest Of Program continues here...

Routine to generate a table of screen addresses

GENTAB	LD DE,16384
		LD HL,SCRADTAB
		LD B,192

LINELOOP	LD (HL),E
		INC H
		LD (HL),D
		DEC H
		INC L

		INC D
		LD A,D
		AND 7
		JR NZ,NEXTLINE
		LD A,E
		ADD A,32
		LD E,A
		JR C,NEXTLINE
		LD A,D
		SUB 8
		LD D,A

NEXTLINE	DJNZ LINELOOP

		RET

Sprite Graphics Data

		ORG 65368

SPRGFX	DEFB %00000000
		DEFB %01111000
		DEFB %01110000
		DEFB %01111000
		DEFB %01011100
		DEFB %00001110
		DEFB %00000100
		DEFB %00000000

OR, XOR, AND

00111100  OR 	00011000  = 00111100

00111100  XOR	00011000  = 00100100

00111100  AND	00011000  = 00011000

Masked Sprites

Masking is a technique used to blank out the area behind a sprite before drawing the sprite to the display. It’s slower than planar (non-masked) sprites, and uses twice as much memory but the final result is more pleasing to the eye. If your sprites are going to be moving over a patterned background its the best technique to use.

The way to draw a masked sprite is to AND the mask with background and then OR the actual sprite with the result. When you AND two bitmaps, bits in the BACKGROUND are reset by zeros in the MASK, and left as they are by ones. That’s probably not very easy to grasp so look at the following examples:

background	            mask

  00111100     and      11111111 	=  00111100	; each bit in the data byte remains  the same after the mask is applied

  00111100     and      00000000 	=  00000000	; all the bits in the data byte are reset by the mask

  11111111     and      00111100 	=  00111100	; the middle four bits remain unchanged and the rest are reset

The previous example showed how to code a very simple sprite routine. Lets now look at a more complex example. This routine will draw a 16 x 16 Masked Sprite, storing the background underneath in a buffer.

This is probably the largest size that can be conveniently shifted in real-time. Any larger and you will want to use a different technique.

The following routine uses a couple of non-standard programming techniques. These tricks help to speed up execution but make the code a bit more difficult to follow. One technique involves using the Stack to retrieve the sprite data, the other is self-modifying code, where the program pokes values into instructions that will be executed later on in the routine.

		ORG 50000,50000

SPRITE	DI				; Disable Interupts.
		LD (SPTEMP),SP		; Store Stack Pointer as it will be redirected
						; by the routine
		LD BC,(XPOS)
		LD A,C
		AND 7
		LD (BITPOS),A

		LD H,SCRADTAB/256		; Calculate screen address as before
		LD L,B
		LD A,(HL)
		INC H
		LD H,(HL)
		LD L,A
		LD A,C
		AND 248
		RRCA
		RRCA
		RRCA
		ADD A,L
		LD L,A

		LD (SCRADD),HL

		LD IX,BKGRNDBUF		; IX is used to point to a temporary buffer where the contents
						; of the screen under the sprite are stored.

		LD SP,SPRGFX		; start address for the sprite graphic

		LD A,16			; The sprite is 16 lines tall
		LD (LINECOUNT),A

LINELOOP	POP DE			; Using the stack to retrieve a line of sprite graphics
		POP HL			; E and D will contain the MASK as it is stored first
						; L and H hold the actual sprite image
		LD A,255			; When shifting, the A reg will be the extra register that the MASK
						; is shifted into. It is loaded with 255 as that means all the bits are
						; initially transparent.
		SCF				; The carry flag is set by this instruction so that bits shifted into the
						; left side of the mask are set (transparent)
		EX AF,AF'			; We also need an extra register for the sprite image, so we will use
						; the alternate or shadow A register.

		LD A,(BITPOS)			; Retrieve the pixel position
		OR A				; Quick way of testing if A is zero
		JR Z,SKIPROTATE		; If zero then no shifting is needed

		LD B,A				; loop counter for number of times to shift

		XOR A				; clear carry flag and set A reg. to zero,
						; as the rotate instructions will shift its contents (carry flag's)
						; into the sprite image data. This is the opposite to the MASK.

ROTATELOOP	EX AF,AF'			; Shift the Sprite Mask data
		RR E				
		RR D
		RRA

		EX AF,AF'			; Shift the Sprite Image data
		RR L
		RR H
		RRA

		DJNZ ROTATELOOP		; Loop back until shifting is complete

SKIPROTATE	LD (DATA+1),A			; This is the piece of self-modifying code mentioned earlier.
		EX AF,AF'			; As the A reg will be needed and there are no other registers
		LD (MASK+1),A			; free, we will poke their data (the rightmost bytes of the sprite mask
						; and image) into instructions further on in the routine.

		LD BC,(SCRADD)		; Get back the screen address

		LD A,(BC)			; Actually put a line of sprite data on the screen
		LD (IX),A			; Store what is under the sprite so that it can be erased
						; without corrupting the background
		AND E				; This instruction applies the MASK
		OR L				; The sprite image is then ORed with the background
		LD (BC),A			; and the result copied back to the screen
		INC C
		INC IXL

		LD A,(BC)			; Put the second byte on the screen
		LD (IX),A			; As we are dealing with a 16 x 16 sprite
		AND D				; each line will be three bytes wide when shifted
		OR H
		LD (BC),A
		INC C
		INC IXL

		LD A,(BC)			; Put the third byte on the screen
		LD (IX),A
MASK		AND 255			; These are the two instructions that were modified
DATA		OR 0				; earlier. The data stored in the A reg. and its shadow
						; were poked into the AND & OR instructions.
		LD (BC),A
		INC IXL

		DEC C
		DEC C				; move back and down one line
		INC B

		LD A,B				; It is not necessary to recalculate
		AND 7				; the screen address for each line of
		JR NZ,A1			; the sprite. All that is needed is to
		LD A,C				; check if a char. or segment (screen third) boundary
		ADD A,32			; has been crossed and adjust the address
		LD C,A				; accordingly.
		JR C,A1
		LD A,B
		SUB 8
		LD B,A

A1		LD (SCRADD),BC		; store it again

		LD HL,LINECOUNT
		DEC (HL)
		JP NZ,LINELOOP		; loop back until all lines are drawn.

		LD SP,(SPTEMP)		; Restore the Stack Pointer
		EI				; Enable Interupts again
		RET

In order to clear the sprite we need to copy back to the screen the Buffer holding the contents
of the screen under the sprite. The following routine will do this.

CLEARSPRITE	DI

		LD BC,(OLDXPOS)

		LD H,SCRADTAB/256
		LD L,B
		LD A,(HL)
		INC H
		LD H,(HL)
		LD L,A
		LD A,C
		AND 248
		RRCA
		RRCA
		RRCA
		ADD A,L
		LD L,A

		LD DE, BKGRNDBUF		

		LD C,16	

CLOOP		LD A,(DE)
		LD (HL),A
		INC L
		INC E

		LD A,(DE)
		LD (HL),A
		INC L
		INC E

		LD A,(DE)
		LD (HL),A
		INC E

		DEC L
		DEC L
		INC H

		LD A,H
		AND 7
		JR NZ,A2
		LD A,L
		ADD A,32
		LD L,A
		JR C,A2
		LD A,H
		SUB 8
		LD H,A

A2		DEC C
		JP NZ,CLOOP

		EI
		RET

XPOS		DEFB 0
YPOS		DEFB 0
OLDXPOS	DEFB 0
OLDYPOS	DEFB 0
LINECOUNT	DEFB 0
BITPOS	DEFB 0
SCRADD	DEFW 0
SPTEMP	DEFW 0

SCRADTAB	EQU 64000
BKGRNDBUF	EQU 64512
; ASM source file created by SevenuP v1.12
; SevenuP (C) Copyright 2002-2004 by Jaime Tejedor G¢mez, aka Metalbrain

;GRAPHIC DATA:
;Pixel Size:      ( 16,  16)
;Char Size:       (  2,   2)
;Sort Priorities: X char, Char line, Y char, Mask
;Attributes:      No attributes
;Mask:            Yes, mask before

SPRGFX	DEFB	248, 31,  0,  0
		DEFB	224,  7,  3,192
		DEFB	192,  3, 15,240
		DEFB	128,  1, 25,248
		DEFB	128,  1, 51,252
		DEFB	  0,  0, 39,252
		DEFB	  0,  0,111,250
		DEFB	  0,  0,127,252

		DEFB	  0,  0,127,250
		DEFB	  0,  0,127,252
		DEFB	  0,  0, 63,248
		DEFB	128,  1, 63,244
		DEFB	128,  1, 31,232
		DEFB	192,  3, 13, 80
		DEFB	224,  7,  2,128
		DEFB	248, 31,  0,  0

Addendum: There appears to be a minor bug in the original code by Derek Smith. Commentator uglifruit posted a fix in the comments section, which has now been integrated into the above code.

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Categories: Z80 Assembly Tags: , ,
  1. uglifruit
    July 12, 2011 at 11:06 pm

    Hi there, came across your (reposting) of the sprite handling piece. I’ve decided to start a new speccy game, after doing some bedroom programming about 20 years ago – so this was a nice refresher.

    There is a slight omission in your code that stops it functioning quite as it should – it is a missing label in the CLEARSPRITE routine:

    The JR, A1 (if left as they are) compile to jump into the SPRITE routine, whereas I am pretty sure they should jump to the DEC C instruction in the later within CLOOP.

    I suggest the following (amended) routine, which fix the problem:

    CLEARSPRITE DI

    LD BC,(OLDXPOS)

    LD H,SCRADTAB/256
    LD L,B
    LD A,(HL)
    INC H
    LD H,(HL)
    LD L,A
    LD A,C
    AND 248
    RRCA
    RRCA
    RRCA
    ADD A,L
    LD L,A

    LD DE, BKGRNDBUF

    LD C,16

    CLOOP LD A,(DE)
    LD (HL),A
    INC L
    INC E

    LD A,(DE)
    LD (HL),A
    INC L
    INC E

    LD A,(DE)
    LD (HL),A
    INC E

    DEC L
    DEC L
    INC H

    LD A,H
    AND 7
    JR NZ,A2 ;notice – this is no longer A1 which is within the Sprite Printing routine
    LD A,L
    ADD A,32
    LD L,A
    JR C,A2 ;notice – no longer A1
    LD A,H
    SUB 8
    LD H,A
    A2 ;using A2 label for symmetry with the Sprite Printing routine
    DEC C
    JP NZ,CLOOP

    EI
    RET

    XPOS DEFB 0
    YPOS DEFB 0
    OLDXPOS DEFB 0
    OLDYPOS DEFB 0
    LINECOUNT DEFB 0
    BITPOS DEFB 0
    SCRADD DEFW 0
    SPTEMP DEFW 0

    SCRADTAB EQU 64000
    BKGRNDBUF EQU 64512

    I hope that helps someone, I had a difficult couple of hours trying to figure why that wasn’t working as I expected!

    All the best, keep up the good work.

  2. uglifruit
    July 12, 2011 at 11:08 pm

    Damn, copy and paste buggered up the tabbed formatting!

    JR NZ,A2

    JR C,A2

    and adding the label A2 before the DEC C

    are the changes!

    • August 10, 2012 at 12:11 pm

      Thanks for the bug-fix. I have updated the code accordingly.

  3. Adler
    August 9, 2012 at 8:03 pm

    Somehow it does not nothing at all when I run it in emulator…any ideas ?

  4. Adler
    August 9, 2012 at 8:13 pm

    I found the problem:
    LD (HL),A ; this always start wriing to adress 0x0000, then each step writing +0x0010

    This is not tested at all ?..Or what ?

    • August 10, 2012 at 12:16 pm

      Which part of the code are you trying to run? Have you set up the addresses correctly?

      • Adler
        August 10, 2012 at 4:31 pm

        I ran the code with the self-modifying part…

        I guess it would be best to put the example in some emulator file format(sna, tap) to be able to run it without any complications. Is that able for you ?

  5. RDK
    July 2, 2014 at 6:04 am

    Hi. I know by this time the post must be a bit old but I beg you to consider this.

    I might be wrong but the mnemonic for LD (IX),A should read LD(IX+0),A as my assemblers won’t compile if that’s written like that.

    • July 3, 2014 at 1:53 pm

      My understanding is that LD (IX), A is a common short hand usage for LD (IX + 0), A. Besides, I didn’t want to change the original code too much as they are being reproduced with the permission from the author. :)

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