; Pointer: puts a joystick-controlled pointer on the screen. ************************ * Set up the screen * ************************ section hello,code_c include source:include/hardware.i move.l 4,a6 Usual stuff for setting up a 3 bitplane screen lea library,a1 jsr -408(a6) move.l d0,base move.l #40*256*3,d0 move.l #2,d1 jsr -198(a6) move.l d0,b1 move.w d0,b1l swap d0 move.w d0,b1h swap d0 add.l #40*256,d0 move.l d0,b2 move.w d0,b2l swap d0 move.w d0,b2h swap d0 add.l #40*256,d0 move.l d0,b3 move.w d0,b3l swap d0 move.w d0,b3h move.l #pointer,d0 This is the pointer, sprite 0 move.w d0,s0l swap d0 move.w d0,s0h lea $dff000,a5 move.l #copper,COP1LCH(a5) move.w #0,COPJMP1(a5) move.l b1,a1 move.l #30*256,d0 clear move.l #0,(a1)+ Clear screen dbra d0,clear clr.l xpos move.l #$29483400,pointer Put pointer at start (if it's rerun, the sprite pointer needs to be put back at top left hand corner) move.l $6c,oldint+2 Get my interrupt in move.l #inter,$6c ************************* * Waits for LMB or FB * ************************* wait btst #7,$bfe001 Has fire button been pressed beq.s fire If so, do routine btst #6,$bfe001 Is LMB pressed bne.s wait No, then loop again ************** * Clean up * ************** move.l oldint+2,$6c move.l base,a2 move.l 38(a2),COP1LCH(a5) move.l #40*256*3,d0 move.l b1,a1 jmp -210(a6) *********************************** * Executed when fire is pressed * *********************************** ; See explanation of how it works after source first ; At the moment the byte the pointer is pointing to is made into 255 ; (all bits set) fire clr.l d0 get ypos into a `clean' d0 move.w ypos,d0 mulu #40,d0 x 40 (40 bytes per line) move.w xpos,d1 get xpos lsr.w #2,d1 / 4 (2*2) since xpos counts 1 per 2 bytes (see below) add.w d1,d0 add xpos/4 to d0 add.l b1,d0 add start of 1st screen move.l d0,a1 Now d0 pointing to wanted byte. Put d0 into a1 move.b #-1,(a1) so it can be used to make the byte 255 bra wait and loop back ; At the moment, the fire button is continually registered when pressed. ; If you wait the fire button to be released, it means the fire button ; will only be registered once per press. ***************** * Copper List * ***************** ; much as you'd expect, except for a couple of bits even copper dc.w BPLCON0,%11001000000000 dc.w BPLCON1,0 dc.w BPLCON2,$20 Doing this means that the sprite has priority over the bitplane (making it 0 means the pointer gets hidden when it goes onto a byte that is more than 0) dc.w BPL1MOD,0 dc.w BPL2MOD,0 dc.w DIWSTRT,$2981 dc.w DIWSTOP,$29c1 dc.w DDFSTRT,$3d dc.w DDFSTOP,$d0 dc.w BPL1PTL b1l dc.w 0,BPL1PTH b1h dc.w 0,BPL2PTL b2l dc.w 0,BPL2PTH b2h dc.w 0,BPL3PTL b3l dc.w 0,BPL3PTH b3h dc.w 0,SPR0PTL s0l dc.w 0,SPR0PTH s0h dc.w 0 dc.w COLOR17,$c00 Red dc.w COLOR18,$cc0 Yellow dc.w COLOR19,0 Black dc.w $ffff,$fffe ; Program variables b1 dc.l 0 b2 dc.l 0 b3 dc.l 0 base dc.l 0 library dc.b 'graphics.library',0,0 ********************************* * The joystick pointer sprite * ********************************* pointer dc.w $2948,$3400 dc.w %0111110000000000,%0111110000000000 dc.w %1000001000000000,%1111111000000000 dc.w %1111010000000000,%1000110000000000 dc.w %1111101000000000,%1000011000000000 dc.w %1111110100000000,%1001001100000000 dc.w %1110111010000000,%1010100110000000 dc.w %0100011101000000,%0100010011000000 dc.w %0000001110100000,%0000001001100000 dc.w %0000000111100000,%0000000100100000 dc.w %0000000011000000,%0000000011000000 dc.w %0000000000000000,%0000000000000000 dc.w 0,0 xpos dc.w 0 ypos dc.w 0 *********************** * The Pointer Mover * *********************** ; The way the program tests the joystick is exactly the same as my ; `Move_Sprite' program which you may still have but if you're ; sensible you'll have written over. If you still don't understand ; it, an explanation is at the very end. It isn't here cos I didn't ; want to patronise you if did understand. ; Some general notes on this part first: ; Since the pointer needs to move over all the screen, the Most ; Significant Bit for both the x position and the y positions will ; need to be changed by the program (remember, these bits are stored ; in the second control word on their own rather than part of the ; actual number: ; First control word: ; bits 8-15 lower 8 bits of VPOS start and bits 0-7 lower 8 bits of HPOS ; Second control word: ; bits 8-15 lower 8 bits of HPOS end, bit 0 highest bit of HPOS ; bit 1 highest bit of VPOS end and bit 2 highest bit of VPOS start) ; That must be the longest bracketed section I've ever done. Since ; all three of these values must be able to go to more than 255, these ; highest bits must be changed. The Carry Flag is used below because ; it is set if either: ; 1. 1 is subtracted from a byte when it is 0 (the byte becoming 255) ; 2. 1 is added to a byte when it is 255 (the byte becoming 0) ; otherwise it is clear. This can be used to see if the highest bit ; must be changed (eg. if a number is going from 255 to 256, the lowest ; 8 bits will be 0, the carry will be set, and the highest bit must ; be set. This is important ; Because this way of storing the numbers is hard to use, the variables ; xpos and ypos have been used as flags to tell where abouts the pointer ; is. For some reason, the pointer moves 2 bits right or left, but ; 1 bit up or down. This means that ypos does hold the normal line ; line number of the screen (0-255), but that since xpos is only altered ; by 1 when the pointer moves 2, xpos is only half of the actual ; horizontal position. This is really easy to change if it fusses you ; (I couldn't care less). inter movem.l d0-d7/a0-a6,-(sp) Yep - save all registers move.b $dff00c,d0 get left/up movement for joystick beq.s nolu no l/u movement if it =0, so skip all l/u routine btst #1,d0 bit 1 is set if it is going left or left and up beq.s up so if its clear, test for up tst xpos is xpos already 0? beq.s up Yes, then it can't move any further left subq.w #1,xpos take 1 from xpos subq.b #1,pointer+1 subtract 1 from pointer's x position bcc.s up if new x position is not 255, go to up and.b #$fe,pointer+3 if it is, the highest HPOS bit becomes 0 up subq.b #1,d0 take 1 from joystick position btst #1,d0 now bit 1 is 0 if it is going up or up and left bne.s nolu if it ain't, go to right/down movement tst ypos is it at the top already? beq.s nolu Yes, then don't move it subq.w #1,ypos take 1 from ypos subq.b #1,pointer move top of pointer up bcc.s ncarry if new number is not 255, go to nolu and.b #$fb,pointer+3 if it is, the highest VPOS start bit becomes 0, (it was 256 before, so this bit would be set) ncarry subq.b #1,pointer+2 move bottom of pointer up bcc.s nolu again, if it ain't 255, go to nolu and.b #$fd,pointer+3 It is 255, so highest bit if VPOS end bit must be cleared nolu move.b $dff00d,d0 Get right/down position beq.s over if its 0, there's no movement so go away btst #1,d0 bit 1 is set if joystick going right or right and down beq.s down so if its clear, test for down cmp.w #159,xpos has it reached far right? (320/2-1) beq.s down if so, don't move addq.w #1,xpos add 1 to xpos addq.b #1,pointer+1 add 1 to pointer number bcc.s down if new number not 256, go to down or.b #1,pointer+3 if it is, highest bit of HPOS must be set down subq.b #1,d0 take 1 from r/d position btst #1,d0 is bit 1 clear bne.s over No, then not going down, so quit cmp.w #255,ypos is it at the bottom already? beq.s over yes, then quit addq.w #1,ypos add 1 to ypos addq.b #1,pointer add 1 to pointer y pos bcc.s ncarry1 if no carry, don't make highest bit=1 or.b #4,pointer+3 if carry, do ncarry1 addq.b #1,pointer+2 same for VPOS end bcc.s over or.b #2,pointer+3 over movem.l (sp)+,d0-d7/a0-a6 old registers oldint jmp $0 old interrupt ; Sorry if I've either gone into too much detail where you understood ; it anyway or not enough where you didn't. ; THE JOYSTICK ; $dff00c ; bit 1 0 | Direction ; ---------+---------- ; 0 0 | neither left nor up ; 0 1 | left only ; 1 0 | up and left ; 1 1 | up only ; as you can see, bit 1 is set if it is going up. If you check for ; no movement (neither bit set) first, and then check for upward ; movement by testing bit 1 for being set. Then subtract 1 from ; the value. left only becomes 00, up and left 01 and up only 10 ; Now, bit 1 is clear if there is leftward movement ; $dff00d ; 0 0 | Neither right nor down ; 0 1 | right only ; 1 0 | down and right ; 1 1 | down only ; As you can see, its just the same for right and down. Again, if ; you already knew all this, you've been wasting you time!