我试图用汇编语言编写有限状态机，但我卡住了

Question

有许多有限状态机提出的问题，但都与我的问题无关。

我需要5种方法

S0     S1   S2   S3   and read the input

我们从里面开始

S0

我们想打印状态→ 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

这是我的代码：这里的问题是输入不起作用。如果我输入00,01,11-

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

Answer 1

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

Answer 2

我假设a和b是你的状态？所以你把状态存储在那里，但你调用函数在两者之间，所以我假设ebx在你检查它之前被丢弃了。

call writestring
call readint
mov a,ebx

mov edx,offset prompt2
call writestring
call readint
mov b,eax

因此，在这里，您至少需要将ebx恢复到，然后才能进行检查（eax已经包含该值）。

mov  a, ebx

不确定是否应该在eax中，因此您可能还必须交换它们。

 xchg eax, ebx

另外，我有点惊讶你调用读音并将ebx移动到a，然后在你再次调用读音之后，但这次将eax移动到b。我认为 readint 返回 eax 中的值，对吧（您没有提供代码）？那么，在第一次调用时，ebx 的值是多少呢？它可能也应该是

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