Assume I want to multiply a large number by another (maybe small) number in assembly. The big number (multiplicand) is saved in DX:AX and the multiplier is saved in BX. The MUL instruction only operates on AX. So what to do with DX?
For example, the number is 0001:0000H (65536) and I want to multiply it by 2.
number dw 0000h, 0001h
...
mov ax, [number]
mov dx, [number+2]
mov bx, 2
mul bx ; it is ax*2 or 0000*2
Therefore the result is zero! Any idea on that?
Let's pretend this is 286, so you don't have eax.
number dd 0x12345678 ; = dw 0x5678, 0x1234
result dw 0, 0, 0 ; 32b * 16b = 48b needed
...
mov ax,[number] ; 0x5678
mov cx,[number+2] ; 0x1234 ; cx, dx will be used later
mov bx,0x9ABC
; now you want unsigned 0x12345678 * 0x9ABC (= 0xB00DA73B020)
mul bx ; dx:ax = 0x5678 * 0x9ABC
; ^ check instruction reference guide why "dx:ax"!
xchg cx,ax
mov di,dx ; di:cx = intermediate result
mul bx ; dx:ax = 0x1234 * 0x9ABC
; put the intermediate multiplication results together
; into one 48b number dx:di:cx
add di,ax
adc dx,0
; notice how I added the new result as *65536 to old result
; by using different 16bit registers
; store the result
mov [result],cx
mov [result+2],di
mov [result+4],dx
It's the same way as when you multiply numbers on paper, just you don't move by *10 components, but exploit the 16b register size nature to move by *65536 (0x10000) components to make it in less steps.
I.e.
13
* 37
----
91 (13 * 7)
39_ (13 * 3, shifted left by *base (=10))
---- (summing the intermediate results, the 39 "shifted")
481 (13 * 37)