Question

仅当此算法最常见时，我的程序才会显示此显示：

digits1 digits2 digits3%DIG == digits2 digits3%DIG == digits3%DIG = rem

dig1,2,3 = the digits of the nember
Dig = the Divisor

该计划非常有效，（000,1）; （999,1）... 但如果我像这样尝试输入（999,5）; （999,9）该程序返回＆＃34;分区再保留不等于＆＃34; 当正确的回归应该是：

"999%5 = 99%5 = 9%5 = 4

感谢您的帮助

;
;  LAB 3 KFIR & ILAN
;
    .MODEL SMALL
    .STACK 100h
    .DATA
DisplayString DB 'Please Enter Number (from 000 up to 999):',13,10,'$'  
DisplayStringDIG DB 13,10,'Please Enter Digit (from 0 up to 9):',13,10,'$' 
DisplayStringResY DB 13,10,'XXX mod X == XX mod X == X mod X = X',13,10,'$'
DisplayStringResN DB 13,10,'The divisions remainer is NOT equal',13,10,'$'

d1 DB ?
d2 DB ?
d3 DB ?
dig DB ?
NUM DB ?
rem DB ? 
ten db 10  
handret DB 100


     .CODE
ProgStart:
     MOV AX,@DATA             ; DS can be written to only through a register
     MOV DS,AX                ; Set DS to point to data segment

     MOV AH, 9              ; set print option for int 21h
     MOV DX, OFFSET DisplayString       ; set DS:DX to point to DisplayStirng  
     INT 21h    


     MOV AH, 1      ; set read option for int 21h
     INT 21h        ; read     
     MOV d1 ,AL 
     MOV DisplayStringResY[2],AL ; Put insert user digit in the original string                                                    
     SUB d1, '0'   ; Ascii to int 

     MOV AH, 1      ; set read option for int 21h
     INT 21h        ; read     
     MOV d2 ,AL
     MOV DisplayStringResY[3],AL; Put insert user digit in the original string
     MOV DisplayStringResY[15],AL                                                   
     SUB d2, '0'   ; Ascii to int 

     MOV AH, 1      ; set read option for int 21h
     INT 21h        ; read     
     MOV d3 ,AL 
     MOV DisplayStringResY[4],AL; Put insert user digit in the original string 
     MOV DisplayStringResY[16],AL
     MOV DisplayStringResY[27],AL                                                     
     SUB d3, '0'; Ascii to int 

     MOV AH, 9              ; set print option for int 21h
     MOV DX, OFFSET DisplayStringDIG ;  
     INT 21h    

     MOV AH, 1      ; set read option for int 21h
     INT 21h        ; read     
     MOV dig ,AL 
     MOV DisplayStringResY[10],AL; Put insert user digit in the original string
     MOV DisplayStringResY[22],AL
     MOV DisplayStringResY[33],AL                                                   
     SUB dig, '0'

     MOV AL ,d3
     MOV AH,0
     DIV dig
     MOV rem, ah 
     ADD AH,'0'  ; int to Ascii
     MOV DisplayStringResY[37],AH; Put insert rem digit in the original string


     mov al,d2
     mul ten
     add al,d3
     MOV NUM,AL

     MOV AH,0
     DIV dig
     CMP AH,rem     
     JNE WorngRES ; Jump and print mess if not Equal     

     MOV AL,d1  
     mul handret
     add AL,NUM          

     MOV AH,0          
     DIV dig
     CMP AH,rem     
     JNE WorngRES  ; Jump and print mess if not Equal


     MOV AH, 9                              ; set print option for int 21h
     MOV DX, OFFSET DisplayStringResY       ; set DS:DX to point to DisplayStirng  
     INT 21h          
     JMP EXIT

WorngRES:
     MOV AH, 9              ; set print option for int 21h
     MOV DX, OFFSET DisplayStringResN       ; set DS:DX to point to DisplayStirng  
     INT 21h              



EXIT:

     mov ah,4ch        
     INT 21h                  ; Return to DOS (terminate program)
    END ProgStart

ASM：分歧溢出

0 个答案: