16 位 x 32 位的汇编乘法 => 48 位

Question

假设我想在汇编中将一个大数乘以另一个（可能很小）的数。大数（被乘数）DX:AX保存在BX. 该MUL指令仅对 进行操作AX。那么该怎么办DX呢？

例如，数字是0001:0000H(65536)，我想将其乘以 2。

number     dw   0000h, 0001h
...
mov    ax, [number]
mov    dx, [number+2]
mov    bx, 2
mul    bx   ; it is ax*2 or 0000*2

所以结果为零！对此有什么想法吗？

Answer 1

让我们假设这是 286，所以你没有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

这与在纸上乘以数字时的方式相同，只是您不移动 *10 个分量，而是利用 16b 寄存器大小的性质来移动 *65536 (0x10000) 个分量以减少步骤。

IE

  13
* 37
----
  91 (13 * 7)
 39_ (13 * 3, shifted left by *base (=10))
---- (summing the intermediate results, the 39 "shifted")
 481 (13 * 37)