Divide Overflow Error In Assembly Language
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Divide By Zero Or Overflow Error
language up vote 0 down vote favorite I have a simple assembly program, where I want to divide two numbers (two byte sized) and print remainder. Here's my code .model small .stack 256 .data ten dw 10 .code main proc mov ax, @data mov ds, ax mov ax, 12 ; divident div ten ; ax/10 mov ah, 9 ; for printing dx int 21h mov your program caused a divide overflow error ax, 4c00h ; ending program int 21h main endp end main So when I run this code the result is "Divide overflow" and I have no idea why does overflow happens. Any ideas? assembly x86 division share|improve this question edited Apr 6 '13 at 11:52 nrz 7,71721453 asked Apr 6 '13 at 11:34 nabroyan 1,2121526 1 Have you tried searching for assembly division problems/questions? Your question is not the first one. It's a duplicate of others. –Alexey Frunze Apr 6 '13 at 11:50 add a comment| 3 Answers 3 active oldest votes up vote 1 down vote accepted DIV mem16 divides DX:AX by the given operand. So you need to sign-extend AX into DX, which you easily can do with the CWD instruction prior to the division (in your case MOV DX,0 or XOR DX,DX would also work). I.e.: mov ax,12 cwd div ten Another problem with your code is that you seem to assume that int 21h / ah=9 can be used to print numeric values. That is not the case. If you want to print the value of DX (i.e. the remainder) you'll have to convert it into a st
- The Netwide Assembler > NASM Forum > Example Code > program errors : divide overflow « previous next » Pages: [1] Print Author Topic: program errors : divide overflow (Read 4932 times) nobody Guest program errors : divide overflow « on: June 17, 2005, 12:51:29 PM » HI alli'm new to this site, but i'm hoping someone can help. I have a program to write to input 6 numbers division overflow error as ascii characters add them, divide by six and output the result. I have functions that convert to numeric and from numeric
Spinrite Division Overflow Error
back to string, but when i ask for a keyboard input, i get a divide overflow.I know it must be in the num_to_str where i have idiv bx, but i cannot figure out
Divide Overflow In Computer Architecture
why. WHen i don't ask for an input and start with '5' as input, i get the answer 9 on the screen no matter what. below is the code. thank you for any help you can provide. my email address is niri@ananzi.co.zaThanks and kind regards,NIRI-----------------------------------------------------------------bits 16org 0x100jmp maininit_mess: db http://stackoverflow.com/questions/15850409/divide-overflow-in-assembly-language 'Please enter the numbers you selected : ',0ah,0dh,'$'input_buf: db 2outputbuf: db ' ','$'blank: db ' ','$'next_mess: db 'Number ','$'store: db 'Stored value right now is',0ah,0dh,'$'input: db '2','$'line_ret: db 0ah,0dh,'$'display_number: mov dx,next_mess mov ah,09 int 21h retdisplay_lineret: mov dx,line_ret mov ah,09 ; Service - display of stringint 21hretdisplay_colon: mov https://forum.nasm.us/index.php?topic=662.0 ah,02 mov dl,3ah int 21h retdisplay_n1: mov ah,02 mov dl,31h int 21h retdisplay_n2: mov ah,02 mov dl,32h int 21h retdisplay_n3: mov ah,02 mov dl,33h int 21h retdisplay_dx: mov ah,09 ; Service - display of stringint 21hretread_string: mov ah,0ah mov dx,input_buf int 21h ret;{----------------------------- CONVERT STRING TO NUMBER ----------------------------; INPUT = DX; OUTPUT = AXstr_to_num: xor ax,ax ; Initial value of AX = 0 xor bh,bh ; Initial value of BH = 0 mov cx,10 ; To build integer in ax(multiply by 10) mov si,dx ; DX point to the start of input buffernext_char: mov bl,[si] ; Move contents of memory pointed to by Si to BL cmp bl,0Dh ; Is it a carriage return? je finis ; Yes, we are done cmp bl,39h ; Compare for '9' jg error ; Is greater than 9
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when you try to preform division by zero. This is an illegal operation and causes an exception. Posted on 2006-04-17 09:36:25 by Synfire Re: divide overflow From "Art of Assembly"Division by zero and Division Overflow (they're NOT the same thing):You cannot, on the 80x86, simply divide one eight bit value by another. If the denominator is an eight bit value, the numerator must be a sixteen bit value. If you need to divide one unsigned eight bit value by another, you must zero extend the numerator to sixteen bits. You can accomplish this by loading the numerator into the al register and then moving zero into the ah register. Then you can divide ax by the denominator operand to produce the correct result. Failing to zero extend al before executing div may cause the 80x86 to produce incorrect results! When you need to divide two 16 bit unsigned values, you must zero extend the ax register (which contains the numerator) into the dx register. Just load the immediate value zero into the dx register12. If you need to divide one 32 bit value by another, you must zero extend the eax register into edx (by loading a zero into edx) before the division. When dealing with signed integer values, you will need to sign extend al to ax, ax to dx or eax into edx before executing idiv. To do so, use the cbw, cwd, cdq, or movsx instructions. If the H.O. byte or word does not already contain significant bits, then you must sign extend the value in the accumulator (al/ax/eax) before doing the idiv operation. Failure to do so may produce incorrect results. There is one other catch to the 80x86’s divide instructions: you can get a fatal error when using this instruction. First, of course, you can attempt to divide a value by zero. Furthermore, the quotient may be too large to fit into the