'I don't know why the division produces a quotient and a remainder that are wrong
How to fix this one: I don't know why both the quotient and the remainder are wrong.
My DOSBox Code:
.model small
.stack 100h
.data
Dividend db 0dh,0ah,"Enter Dividend : $" ;string
Divisor db 0dh,0ah,"Enter Divisor : $"
Quotient db 0dh,0ah,"Display Quotient : $"
Remainder db 0dh,0ah,"Display Remainder : $"
.code
main proc ;main program here
mov ax,@data ;initialize ds
mov ds,ax
mov ah,09h ;Show Enter Dividend
lea dx, Dividend
int 21h
mov ah,01h ;Input Dividend
int 21h
mov bh,al
mov ah,09h ;Show Enter Divisor
lea dx, Divisor
int 21h
mov ah,01h ;Input Divisor
int 21h
mov bl,al
mov ah,00h ;Divide
mov al,bh
div bl
mov cx,ax
add cx,3030h
mov ah,09h ;Show Display Quotient
lea dx, Quotient
int 21h
mov ah,02 ;Display Quotient
mov dl,cl
int 21h
mov ah,09h ;Show Display Remainder
lea dx, Remainder
int 21h
mov ah,02 ;Display Remainder
mov dl,ch
int 21h
mov ah,4Ch ;end here
int 21h
main endp
end main
Error Result:
Enter Dividend : 7 Enter Divisor : 3 Display Quotient : 1 Display Remainder : 4
Solution 1:[1]
The strange results that you show us stem from the fact that the program executes the division 55 / 51 which will indeed produce a quotient of 1 and a remainder of 4.
The DOS.GetCharacter function 01h that you use for input returns an ASCII code in AL
. When you push 7 on the keyboard, the AL
register will hold the ASCII code 37h or 55 in decimal. Before using this input in a calculation you need to convert from character into the value that the digit represents. A simple sub al, 30h
does that. Of course you need to do the same thing for the second input too:
...
mov ah, 09h ;Show Enter Dividend
lea dx, Dividend
int 21h
mov ah, 01h ;Input Dividend
int 21h
sub al, 30h
mov bh, al
mov ah, 09h ;Show Enter Divisor
lea dx, Divisor
int 21h
mov ah, 01h ;Input Divisor
int 21h
sub al, 30h
mov bl, al
...
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | Sep Roland |