这是emu8086中的程序，使用stepper_motor.exe和led_display.exe。
该计划的目的是：
当我在屏幕上输入数字1时，踏板电机顺时针旋转，这会增加LED的数量，当我进入2步进电机逆时针旋转时，这会减少LED的数量。
mov ax, 0
mov [00000h],ax
c dw 4
r macro data
; this is an example of out instruction.
; it writes values to virtual i/o port
; that controls the stepper-motor.
; c:\emu8086\devices\stepper_motor.exe is on port 7
steps_before_direction_change = 8h
jmp start
start:
mov bx, offset data ; start from clock-wise half-step.
mov si, 0
mov cx, 0 ; step counter
next_step:
; motor sets top bit when it's ready to accept new command
wait: in al, 7
test al, 10000000b
jz wait
mov al, [bx][si]
out 7, al
inc si
cmp si, 4
jb next_step
mov si, 0
inc cx
cmp cx, steps_before_direction_change
jb next_step
ret
endm
r1 macro data
; this is an example of out instruction.
; it writes values to virtual i/o port
; that controls the stepper-motor.
; c:\emu8086\devices\stepper_motor.exe is on port 7
steps_before_direction_change1 = 8h
;#start=stepper_motor.exe#
;#make_bin#
; 32 (decimal)
jmp start1
start1:
mov bx, offset data ; start from clock-wise half-step.
mov si, 0
mov cx, 0 ; step counter
next_step1:
; motor sets top bit when it's ready to accept new command
wait1: in al, 7
test al, 10000000b
jz wait1
mov al, [bx][si]
out 7, al
inc si
cmp si, 4
jb next_step1
mov si, 0
inc cx
cmp cx, steps_before_direction_change1
jb next_step1
ret
endm
s:mov ah,01h
int 21h
cmp al,'1'
je l1
jne l11
l11:cmp al,'2'
je l2
jne l12
l12:jmp s
l1:call init
SUB AX,AX
mov ax,c
inc ax
mov c,ax
out 199, ax
r datcw
l2:call init
SUB AX,AX
mov ax,c
dec ax
mov c,ax
out 199, ax
r1 datccw
jmp s
datccw db 0000_0011b
db 0000_0001b
db 0000_0110b
db 0000_0010b
datcw db 0000_0110b
db 0000_0100b
db 0000_0011b
db 0000_0010b
init proc
#start=led_display.exe#
#make_bin#
name led
#start=stepper_motor.exe#
#make_bin#
name "stepper"
ret
init endp
>
但是当我输入1然后是2时它会陷入循环中，当我输入1然后1时，它无法正常工作。
请指导我如何处理它？</ p>