或者，使用
我正在尝试使用6502微控制器指令集生成两个输出：20Hz方波和30Hz方波。到目前为止，我可以在20Hz波上输出：
%uasm65,title="SQUARES"
org 0200h
lda #1d
sta 0a200h
Main:
;jump to the subroutine Delay and do it
jsr Delay
lda 0a200h
inc Count1
lda Count1
cmp #3d
beq Reset1
jmp Main
Reset1:
lda #0d
sta Count1
lda 0a200h
eor #00000001b
sta 0a200h
jmp Main
Reset2:
jmp Main
Delay:
;Save registers on the stack.
pha
txa
pha
tya
pha
;Change the number that is being loaded into the
; 'A' register in order to change the delay time.
lda #01h
OutLoop:
ldx #04h
InLoop1:
ldy #0ffh
InLoop2:
dey
bne InLoop2
dex
bne InLoop1
sec
sbc #1d
bne OutLoop
;Restore registers from the stack.
pla
tay
pla
tax
pla
rts
Count1:
dbt 0d
Count2:
dbt 0d
end
%/uasm65
根据我的理解，我能做的就是取一个60Hz的方波，用它来获得30Hz和20Hz。如何在不影响端口中其他位的状态的情况下，将20Hz的方波输出到端口a的第5位，将30Hz的方波输出到PortA的第6位？换言之，我如何从60岁变成20岁和30岁？我是否让计数检查为7并递增计数2？任何帮助都将不胜感激。
您需要两个独立的计数器，每个引脚一个
Main:
;jump to the subroutine Delay and do it
jsr Delay
lda 0a200h ; ?? what's this doing here?
inc Count1 ; count1 is for the 20 Hz bit pin
lda Count1
cmp #3d ; 60/20 = 3, so counter1 will have to reach 3
bne Skip1 ; otherwise skip toggling
toggle_pin5:
lda #0d ; reload first Counter
sta Count1
lda 0a200h
eor #00000001b
sta 0a200h
skip1:
inc Count2 ; count2 is for the 30 Hz bit pin
lda Count2
cmp #2d ; 60/30 = 2, so counter2 will have to reach 2
bne Skip2 ; you could also "bne Main" here
toggle_pin6:
lda #0d ; reload 2nd Counter
sta Count2
lda 0a200h
eor #00000010b ; you will want to change this for the correct value to "set bit6 of PortA"
sta 0a200h
skip2:
jmp Main
Reset1: ; not needed anymore
Reset2: ; not needed anymore
Delay: [ ... ]
在循环中，分支到Reset1（或Reset2），跳回Main不是一个好主意，你应该跳过第二个引脚的第二次检查。最好只是在几个指令上进行分支（就像我所做的那样），或者使用JSR/RET:
cmp #3d
bne SkipCall ; counter value NOT reached, so skip "Reset"
jsr Reset
SkipCall:
<...>
Reset:
lda #0d
sta Count1
<...>
ret
根据我对指令集的阅读，这应该可以工作，并且比Tommylee的代码更短。（我以此为起点）。
在asm中，如果使用递减来在结果为零时设置零标志，则向零计数更可取。那么你就不需要单独比较了。这可以将代码大小降到0x1D字节（对于我的第二个版本）。
我假设具有内存操作数的dec仍然根据结果设置标志。除了维基百科，我没有看过任何6502个文档P问题中的代码使用dey/bne，所以我认为这是正确的，并设置了标志。
假设针对更少指令进行优化会更好，那么应该尽量减少延迟循环。也许只是以bne为循环条件的嵌套内存循环递减（所以你循环了2^n次）？除非使用内存会消耗更多电力？
Main:
ldx #3d ; 60/20 = 3: toggle every 3 iterations
; stx Count5 ; Count5 is for the 20 Hz bit wave on pin5
ldy #2d ; 60/30 = 2: toggle every 2 iteration
; sty Count6 ; Count6 is for the 30 Hz bit wave on pin6
; omit the stores: Count5 and Count6 are already initialized.
; lda 0a200h ; start with the initial state of the I/O port
lda #1d ; constant initial state
squarewave_loop:
jsr Delay
; lda 0a200h ; or do this here, so Delay doesn't have to save/restore A
dec Count1
bne skip1 ; toggle when it reaches zero
toggle_pin5:
stx Count5 ; reload first countdown counter
eor #00000001b
skip1:
dec Count2
bne skip2 ; toggle when it reaches zero
toggle_pin6:
sty Count6
eor #00000010b ; FIXME: which bit maps to bit6 of Port A?
skip2:
sta 0a200h ; always store, even if there was no state change
jmp squarewave_loop
Delay: [ ... ]
Count1:
dbt 3d
Count2:
dbt 2d
或者，使用dey/dex作为count1/count2
然后我们不需要任何内存来存储计数器，我假设具有内存操作数的指令具有更长的编码
Main:
ldx #3d ; 60/20 = 3: toggle every 3 iterations
ldy #2d ; 60/30 = 2: toggle every 2 iteration
; lda 0a200h ; start with the initial state of the I/O port
lda #1d ; constant initial state
squarewave_loop:
jsr Delay
; lda 0a200h ; or do this here, so Delay doesn't have to save/restore A
dex
bne skip1
;toggle_pin5: ; runs when 1st down-counter hits zero
ldx #3d ; reload the countdown
eor #00000001b
skip1:
dey
bne skip2
;toggle_pin6: ; runs when 2nd down-counter hits zero
ldy #2d
eor #00000010b ; FIXME: which bit maps to bit6 of Port A?
skip2:
sta 0a200h ; always store, even if there was no state change
jmp squarewave_loop
Delay: [ ... ]
此集合在http://www.masswerk.at/6502/assembler.html，如果我去掉注释并从标签末尾删除:字符。不计算延迟循环，总大小为0x1D字节的代码。