Is it possible to multiply by an immediate with mul in x86 Assembly?

Question

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

Answer 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 (requires 186 for imul-immediate):

mov ax,3
imul ax,2  ; imul is for signed multiplication, but low half is the same
           ; the product is in ax.  dx is not modified

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

Answer 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.