我试图用汇编语言编写有限状态机，但我卡住了
有许多有限的状态机问题，但都与我的问题无关。
我需要5种方法
S0 S1 S2 S3 and read the input
我们开始
S0
我们要打印状态→0和输出0→
读取输入首先在ebx中，第二个将在eax中
. If (ebx ==0&&eax==0)
Call S0
.elseif (ebx==1)&&(eax==1)
Call S1
.else
Call S2
.endif
做完整的程序
这里是我的代码：这里的问题是输入不工作。 如果我输入00,01,11 - >这一切给我相同的输出是不正确的。 我想输入00并呼叫S0，输入11呼叫S1。 这不是我为什么不这样做。 任何人都可以弄清楚。
TITLE finite state machine
INCLUDE Irvine32.inc
E = 13
.data
invalidMsg BYTE 'Ivalid input',0
a DWORD ?
b DWORD ?
count dword ?
prompt1 byte 'Enter 0 or 1: ',0
prompt2 byte 'Enter 0 or 1: ',0
num1 byte 'The output is now 1 ',0
num2 byte 'The ouput is now 0',0
num3 byte 'The state is now 0 ',0
num4 byte 'The state is now 1 ',0
num5 byte 'The state is now 2 ',0
num6 byte 'The state is now 3 ',0
.code
main PROC
call clrscr
mov edx,offset prompt1
call writestring
call readint
mov a,ebx
mov edx,offset prompt2
call writestring
call readint
mov b,eax
.if(ebx ==0 && eax == 0)
call S0
.elseif(ebx == 1 && eax == 1)
call S1
.elseif(ebx == 0 && eax == 1)
call S2
.else
call S3
.endif
exit
main ENDP
S0 proc
mov edx,offset num3
call writestring
call crlf
mov edx,offset num2
call writestring
call readint
ret
S0 endp
S1 proc
mov edx,offset num4
call writestring
call crlf
mov edx,offset num2
call writestring
ret
S1 endp
S2 proc
mov edx,offset num5
call writestring
call crlf
mov edx,offset num1
call writestring
call crlf
ret
S2 endp
S3 proc
mov edx,offset num6
call writestring
call crlf
mov edx,offset num1
call writestring
ret
S3 endp
END main
我假设a和b是你的状态？ 因此，您将状态存储在那里，但是您可以在两者之间调用函数，因此我会假设在检查之前ebx已被删除。
call writestring
call readint
mov a,ebx
mov edx,offset prompt2
call writestring
call readint
mov b,eax
所以在这里你需要至少恢复ebx ，然后才能执行检查（eax已经包含该值）。
mov a, ebx
不知道，如果a应该是在eax虽然如此，你可能不得不更换它们。
xchg eax, ebx
另外我有点惊讶，你打电话给readint并且把ebx移到a ，然后再打电话给readint ，但是这次把eax移到b 。 我会认为readint返回eax的值，对（你没有提供代码）？ 那么在第一次通话时ebx会有什么价值？ 它可能应该也是
mov b, eax
更新
mov edx,offset prompt1
call writestring
call readint
mov a,eax
mov edx,offset prompt2
call writestring
call readint
mov b,eax
mov eax, a
mov ebx, b
TITLE Finite State Machine (Finite.asm)
; This program implements a finite state machine that
; accepts an integer with an optional leading sign.
INCLUDE Irvine32.inc
ENTER_KEY = 13
.data
InvalidInputMsg BYTE "Invalid input",13,10,0
.code
main PROC
call Clrscr
StateA:
call Getnext ; read next char into AL
cmp al,'+' ; leading + sign?
je StateB ; go to State B
cmp al,'-' ; leading - sign?
je StateB ; go to State B
call IsDigit ; ZF = 1 if AL contains a digit
jz StateC ; go to State C
call DisplayErrorMsg ; invalid input found
jmp Quit
StateB:
call Getnext ; read next char into AL
call IsDigit ; ZF = 1 if AL contains a digit
jz StateC
call DisplayErrorMsg ; invalid input found
jmp Quit
StateC:
call Getnext ; read next char into AL
call IsDigit ; ZF = 1 if AL contains a digit
jz StateC
cmp al,ENTER_KEY ; Enter key pressed?
je Quit ; yes: quit
call DisplayErrorMsg ; no: invalid input found
jmp Quit
Quit:
call WaitMsg
call Crlf
exit
main ENDP
;-----------------------------------------------
Getnext PROC
;
; Reads a character from standard input.
; Receives: nothing
; Returns: AL contains the character
;-----------------------------------------------
call ReadChar ; input from keyboard call WriteChar ; echo on screen
ret
Getnext ENDP
;-----------------------------------------------
DisplayErrorMsg PROC
;
; Displays an error message indicating that
; the input stream contains illegal input.
; Receives: nothing.
; Returns: nothing
;-----------------------------------------------
push edx
mov edx,OFFSET InvalidInputMsg
call WriteString
pop edx
ret
DisplayErrorMsg ENDP
END main
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