Division

I guys. New to assembly. I have listed my code below an the code solves the folowing problem. A*B -(A+B)/ (A-B). My code works until i get to the division part. Any suggestions would be very helpful. Thanks for reading!





.386
.MODEL FLAT

ExitProcess PROTO NEAR32 stdcall, dwExitCode:DWORD

INCLUDE io.h ; file for input/output

cr EQU 0dh ; carriage return character
Lf EQU 0ah ; line feed character

.STACK 4096 ; 4096-byte stack is reserved

.DATA ; storage for data reserved
numberA DWORD ?
numberB DWORD ?
promptA BYTE "Enter first number: ", 0
promptB BYTE "Enter second number: ", 0
string BYTE 40 DUP (?)
result BYTE cr, Lf, "The answer is "
answer BYTE 11 DUP (?)
BYTE cr, Lf, 0

.CODE ; beginning of main program code
_start:
output promptA ; prompts for the 1st number
input string, 40 ; reads ASCII characters
atod string ; converts to integer
mov numberA, eax ; stores 1st number (A) in memory

output promptB ; prompts for the 2nd number
input string, 40 ; reads ASCII characters
atod string ; converts to integer
mov numberB, eax ; adds 2nd number to the 1st (A+B)

mov eax, numberA ; puts 1st number in eax register (A)
add eax, numberB ; adds 2nd number to the first (A+B)

mov ebx, numberA ; puts 1st number in ebx register
imul ebx, numberB ; multiplies 2nd and 1st number together
sub ebx, eax ; subtracts (A+B) from (A*B)

mov ecx, numberA ; puts 1st number in ecx register
sub ecx, numberB ; subtracts 2nd number from 1st number (A-B)
cwd ; preparing the divisor
idiv ecx ; dividing by (A-B)




dtoa answer, ebx ; converts to ASCII characters
output result ; outputs result label and answer to problem

INVOKE ExitProcess, 0 ; exits with return code 0

PUBLIC _start ; make entry point public

END ; end of code

"idiv ECX" is a 32-bit operation where the dividend is in EDX:EAX, and the divisor is in ECX.
The previous line, "cwd" converts the 16-bit value in the AX register into an equivalent signed value in the DX:AX register pair, by duplicating the sign bit to the left. I'm not sure this is really useful to you. It seems that EAX is already holding a correct value. You only need to clear the EDX register, with "xor edx, edx".
But, maybe there are other things I have not seen.

Edit: Rather than "cdw" (or my "xor edx, edx"), it is "cdq" which is needed. But I read that they have the same opcode (99), they are giving the same results, because they act on AX/DX or on EAX/EDX depending whether it is a 16- or 32-bit segment, and whether there are any operand-size override prefixes.