问题
I am learning assembly for x86 using DosBox emulator. I am trying to perform multiplication. I do not get how it works. When I write the following code:
mov al, 3
mul 2
I get an error. Although, in the reference I am using, it says in multiplication, it assumes AX is always the place holder, therefore, if I write:
mul, 2
It multiplies al value by 2. But it does not work with me.
When I try the following:
mov al, 3
mul al,2
int 3
I get result 9 in ax. See this picture for clarification:
Another question: Can I multiply using memory location directly? Example:
mov si,100
mul [si],5
回答1:
There's no form of MUL that accepts an immediate operand.
Either do:
mov al,3
mov bl,2
mul bl ; the product is in ax
or:
mov ax,3
imul ax,2 ; imul is for signed multiplication, but that doesn't matter here
; the product is in ax
or:
mov al,3
add al,al ; same thing as multiplying by 2
or:
mov al,3
shl al,1 ; same thing as multiplying by 2
回答2:
Intel manual
The Intel 64 and IA-32 Architectures Software Developer’s Manual - Volume 2 Instruction Set Reference - 325383-056US September 2015
section "MUL - Unsigned Multiply" column Instruction contains only:
MUL r/m8
MUL r/m8*
MUL r/m16
MUL r/m32
MUL r/m64
r/mXX means register or memory: so immediates (immXX) like mul 2 are not allowed in any of the forms: the processor simply does not support that operation.
This also answers the second question: it is possible to multiply by memory:
x: dd 0x12341234
mov eax, 2
mul dword [x]
; eax == 0x24682468
And also shows why things like mul al,2 will not work: there is no form that takes two arguments.
As mentioned by Michael however, imul does have immediate forms like IMUL r32, r/m32, imm32 and many others that mul does not.
来源：https://stackoverflow.com/questions/20499141/is-it-possible-to-multiply-by-an-immediate-with-mul-in-x86-assembly