주어진 텍스트에서 단어를 검색하는 x86 어셈블리 프로그램

Question

텍스트에서 단어를 검색 할 수있는 x86 어셈블리를 사용하여 프로그램을 작성하려고합니다. 단어가 텍스트에 있으면 프로그램에서 사용자에게 알려줍니다. 문자열을 비교하는 데 여전히 문제가 있습니다. 어떤 충고?

.model small 

.stack 200h 

.data 

message1 db "Enter your text here: $" 
text db 150,151 dup(0) 
message2 db 10,13,"Enter the word that you want to find: $" 
find db 20,21 dup(0) 
yesmessage db 10,13,"The word is in the text$" 
nomessage db 10,13,"Sorry the word is not in the text$" 

.code 
Start: 

;Display message and key in strings 
mov ax,seg message1 
mov ds,ax 

mov si,offset text 
mov di,offset find 

mov dx,offset message1 
mov ah,09h 
int 21h 

mov dx,si 
mov ah,0Ah 
int 21h 

mov ax,seg message2 
mov ds,ax 

mov dx,offset message2 
mov ah,09h 
int 21h 

mov dx,di 
mov ah,0Ah 
int 21h 

;compare strings 
mov bx,00 

mov bl,text+1 
mov bh,find+1 

cmp bl,bh 
jne L1 

add si,2 
add di,2 

L2:mov bl,byte ptr[si] 
cmp byte ptr[di],bl 
jne L1 
inc si 
inc di 
cmp byte ptr[di],"$" 
jne L2 
mov ah,09h 
mov dx,offset yesmessage 
int 21h 
L1:mov ah,09h 
mov dx,offset nomessage 
int 21H 

mov ax,4c00h 
int 21h 
end start 

예상 된 결과는 다음과 같아야합니다

예 1 :

 
Enter your text here: He is old 

Enter the word that you want to find: old 

The word is in the text 

예 2 : 코드에서

 
Enter your text here: He is old 

Enter the word that you want to find: young 

Sorry the word is not in the text 

Answer 1

내가 분명 몇 가지 문제를 볼 수 있습니다. 난 당신이 나에게 대한보다 적절한 코딩을 보여 see..could

mov bx,00 ; this instruction is redundant 

mov bl,text+1 
mov bh,find+1 

cmp bl,bh 
jne L1 ; it's quite likely that the strings won't have the same length, 
     ; i.e. that find will be shorter than text. this condition is 
     ; therefore incorrect. it would make more sense to use jl, i.e. 
     ; jumping to the negative print if text is shorter than find. 

mov ah,09h 
mov dx,offset yesmessage 
int 21h 
L1:mov ah,09h 
mov dx,offset nomessage ; you'll be printing both messages in cases where 
int 21H     ; the substring is found, because you don't have 
         ; any jump that skips past it.