I am using 8086 emulator and DOSBOX and MASM.
I know that when we multiply 8-bit with 8-bit, answer will be of 16-bit.
al*(8-bit)=ax
An
Imagine you have to do it in the decimal system, e.g. 37 * 8
. You would calculate and add two terms: 30 * 8
+ 7 * 8
. The first term can be transformed to 3 * 8 * base (10)
.
37 * 8
------
56 (7*8)
24 (3*8)
======
296
You see the place of "base" is empty since this value is always 0.
Now let us change base 10 to base "register":
DX:AX * CX
----------
DX:AX (AX*CX)
DX:AX (DX*CX)
==========
XX:XX:XX
You need two multiplications and three words for the result. Furthermore you have to store the result (DX:AX) from the first multiplication because you need the registers for the second multiplication.
Here's the code for 10! I omit the leftmost word of the result since it isn't needed for 3628800:
.MODEL SMALL
.STACK 1000h
.DATA
Result DD 1 ; 01 00 00 00 (little endian!)
.CODE
main PROC
mov ax, @DATA ; Initialize DS
mov ds, ax
mov cx, 2 ; Start value
L1:
mov ax, WORD PTR Result + 0 ; Low word
mul cx
mov di, dx ; Store high result
mov WORD PTR Result + 0, ax ; Low result won't be changed anymore
mov ax, WORD PTR Result + 2 ; High word
mul cx
add ax, di ; add low result from last mul to low result here
mov WORD PTR Result + 2, ax ; Store it
; adc dx, 0 ; this would store the "highest" word of result (13!)
; mov WORD PTR Result + 4, dx
add cx, 1
cmp cx, 10
jbe L1 ; While CX <= 10
mov ax, 4C00h
int 21h
main ENDP
END main
I am pretty sure there is a duplicate of this somewhere already. Of course it's just elementary maths anyway:
result_low = low(input_low * factor)
result_high = high(input_low * factor) + low(input_high * factor)
Pay attention to saving ax
and dx
as necessary.