x86汇编对16位及更高位的指令操作数进行乘法和除法

Question

I'm rather confused about how the multiply and divide operations work in x86 assembly. For example, the code below doesn't seem too difficult since deals with 8-bit.

8-Bit Multiplication:

; User Input:
; [num1], 20
; [num2] , 15

mov    ax, [num1]    ; moves the 8 bits into AL
mov    bx, [num2]    ; moves the 8 bits into BL

mul    bl            ; product stored in AX

print  ax

But what happens when you want to multiply two 16-bit numbers? How would one multiply two 16 bit numbers the same way as it has been done with the 8 bit numbers?

I'm confused as to what registers the values would be stored in. Would they be stored in AL and AH or would it simply store the 16-bit number in AX. To show what I mean:

; User Input:
; [num1], 20
; [num2], 15

mov    eax, [num1]    ; Does this store the 16-bit number in AL and AH or just in AX
mov    ebx, [num2]    ; Does this store the 16-bit number in BL and BH or just in BX

mul    ???            ; this register relies on where the 16-bit numbers are stored

print  eax

Could someone elaborate a bit on how the multiplying and dividing works? (specifically with 16-bit and 32-bit numbers? Would I need to rotate bits if the values are stored in the lower AL and AH?

Or can one simply mov num1 and num2 into ax and bx respectively and then multiply them to get the product in eax ?

Answer 1

A quick glance at the documentation shows that there are 4 possible operand sizes for MUL . The inputs and outputs are summarized in a handy table:

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| Operand Size | Source 1 | Source 2   | Destination |
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| Byte         | AL       | r/m8       | AX          |
| Word         | AX       | r/m16      | DX:AX       |
| Doubleword   | EAX      | r/m32      | EDX:EAX     |
| Quadword     | RAX      | r/m64      | RDX:RAX     |
------------------------------------------------------