There are many finite state machine asked questions but all are not related to my problem.
I need 5 methods
S0 S1 S2 S3 and read the input
We start in
S0
We want to print the state → 0 and the output 0→
Read input First is in ebx and the second will be in eax
. If (ebx ==0&&eax==0)
Call S0
.elseif (ebx==1)&&(eax==1)
Call S1
.else
Call S2
.endif
Do the complete program
here is my code: The problem here is input is not working. If i input 00,01,11 -> it all give me same output which is not right. I want to enter 00 and call S0, enter 11 call S1. It is not doing that i don't why. Can anyone figure out.
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
I assume that a and b are your states? So you store the state there but you call functions in between, so I would assume that ebx is trashed before you check it.
call writestring
call readint
mov a,ebx
mov edx,offset prompt2
call writestring
call readint
mov b,eax
So here you would need to restore at least ebx to before you can do the check (eax already contains the value).
mov a, ebx
Not sure if a is supposed to be in eax though, so you may have to exchange them as well.
xchg eax, ebx
Also I'm a bit surprised that you call readint and move ebx to a and right after you call readint again, but this time move eax to b. I would think that readint returns the value in eax, right (you didn't provide the code)? So what value would be in ebx on the first call? It probably should be also
mov b, eax
update
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