彙編代碼----查找匹配字符串

彙編實驗三代碼，查找匹配字符串，

;*****************************************
datasg  segment para    'data'  
  string1   db  'Enter keyword:$'
  string2   db  'Enter sentence:$'
  temp db  13,10,'$'
  num db ?,13,10,'$'
;
  mess1   db  'Match at location:',?,?,' H of the sentence',13,10,'$'
  mess2   db    'No match.',13,10,'$'  
;定義相關變量
keyword     label   byte
    maxk     db      5
    actk     db      ?           
    keyd   db      5 dup(?)    
sentence     label   byte
    maxs     db      10
    acts     db      ?           
    sente   db      10 dup(?)   
datasg  ends
;*****************************************
codesg  segment para    'code'
        assume  cs:codesg,ds:datasg,es:datasg
;-----------------------------------------
main    proc    far
        push    ds
        sub     ax,ax
        push    ax
        
        mov     ax,datasg
        mov     ds,ax
        mov     es,ax
;main part of program goes here
start:
        lea     dx,string1 ;提示輸入關鍵字
        mov     ah,09
        int     21h
        lea     dx,keyword 
        mov     ah,0ah  
        int     21h
        cmp     actk,0
        je      exit

        lea     dx,temp ;換行，防止覆蓋
        mov     ah,09
        int     21h

        lea     dx,string2 ;提示輸入句子
        mov     ah,09
        int     21h
        lea     dx,sentence
        mov     ah,0ah  
        int     21h
        cmp     acts,0
        je      exit

        lea     dx,temp ;換行，防止覆蓋
        mov     ah,09
        int     21h





        lea     si,keyd
        lea     bx,sente
        mov     di,bx
        cld
        mov     ch,0
        mov     cl,acts  ;句子長度，一個字節
        sub     cl,actk
        inc     cx ;比較次數

comp:
        push    cx
        push    di
        mov     cl,actk ;關鍵字長度，一個字節
        repz    cmpsb  ;如果相等繼續比較;64
        jz      find
        pop     di
        pop     cx
        inc     di
        lea     si,keyd
        loop    comp  ;

        lea     dx,mess2 ;不匹配
        mov     ah,09
        int     21h
        jmp     exit


find:   
        pop     di
        pop     cx
        sub     di,bx
        mov     bx,di
        inc     bx ;得到十進制的位置

        mov     ax,bx
        mov     bh,16
        div     bh
        add     al,48
        add     ah,48
        mov     [mess1+18],al
        mov     [mess1+19],ah


        lea     dx,mess1 ;輸出
        mov     ah,09
        int     21h
        jmp     start


exit:
    ret                 ;return to DOS
main    endp
;-------------------------------------------
codesg  ends            ;end of segment
;*******************************************
        end     main    ;end assembly