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This is the first unoptimized version of the code. You will see the sprite data is sitting at address $4000, and is terminated at sp_end. The byte values are in hex: $00,$00,$A0,$0a,$a4,$aa. The number in the comment field is the cycle count, or the number of clock cycles it takes to execute the instruction. There is a way to get the count via lwasm, but it beats me how to do it on the command line. The backgnd is at address $4100 and finally the destination mixbuf is at $4200. mixbuf is what you look at once the program has finished.
01 org $3f00
02
03 [3] ldu #mixbuf ;buffer 3
04 [3] ldx #sprite ;pointer to sprite stream
05 [4] ldy #backgnd
06
07 [4+0] again lda ,x 4;get sprite byte
08 [5] bne nxt 3;not 0, so we have to do something
09
10 rezero
11 [4+1] leax 1,x 5;was zero, inc to next byte
12 [4+2] ldb ,y+ 6;get bgnd and inc y
13 [4+2] stb ,u+ 6;store background and inc u
14
15 [4+0] lda ,x 4;get next byte
16 [5] beq rezero 3
17
18 [2] nxt bita #$0f 2;lets check if bits are
19 [5] beq poo 3;set in upper and lower
20 [2] bita #$f0 2;half... if either half
21 [5] beq poo 3;is set, we need to do
22 [5] bra copy 3;stuff
23
24 [4+2] poo ldb ,y+ 6;get background byte + inc
25 [2] anda #$0f 2;check if high bits set
26 [5] bne high_set 3
27
28 [4+0] lda ,x 4
29 [2] anda #$f0 2
30 [2] andb #$0f 2
31 [5] bra low_set 3
32
33 [2] high_set andb #$f0 2
34 [5] low_set stb dd+1 5
35
36 [2] dd ora #0 2
37
38 [4+2] copy sta ,u+ 6
39 [4+1] leax 1,x 5
40 [4] spend cmpx #sp_end 4
41 [5] blo again 3
42 [5] render rts
43
44 org $4000
45 sprite fcb $00,$00,$A0,$0a,$a4,$aa
46 sp_end
47 org $4100
48 backgnd fcb $CD,$dd,$dd,$ad,$bd,$cd
49
50 org $4200
51 mixbuf fcb $00,$00,$00,$00,$00,$00
52 ;expected result $cd,$dd,$ad,$aa,$a4,$aa
53
The source above is the cycle count. So lets get to optimizing!
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