x86 assembly multiply and divide instruction operands, 16-bit and higher

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?

Solution

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:

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

Comments

[StartingGroovy]

I looked at that, and was aware of where the value is stored after the mul. I'm wondering if, when I store the value in eax (prior to mul) if it stores it in the lower bits (AL and AH) rather than the higher bits (AX). If it does store them in the lower bits, am I able to rotate them to AX and then just multiply AX and BX and print out the answer (I believe it would be in EAX)?

[Carl Norum]

If you store a value in eax it's in all of eax. I'm not sure I understand. I think you might be confusing which parts of eax are ah, al, and ax, too. You can see the chart in volume 1 of the same documentation. In particular, ax refers to the same 16 bits as ah:al, not the "upper" 16-bits of eax.

[StartingGroovy]

Ah you are correct, that is what I was confused about. So essentially what I would want to do is multiply AX by BX and my product should be in EAX?

[Carl Norum]

Close, but the product will be in DX:AX. Check the table in my answer.

[StartingGroovy]

So multiplying two 16-bit numbers doesn't make a 32-bit number? (This would also be why I was confused with the chart. I thought multiplying two 16-bit numbers would make a 32-bit number)

[Carl Norum]

Sure it does - DX is 16 bits, and AX is 16 bits. That makes DX:AX a 32-bit number. 16+16=32, right?

[StartingGroovy]

Wow, sometimes I think I'm half blind when trying to read documentation. I assumed that meant it stored the values in AX OR DX not both. Last question before I accept your answer: So if I want to print this out, would I move DX into EAX rotate left and then move AX in as well and print EAX?

[Carl Norum]

AX is already there - it's the bottom 16 bits of EAX. You would need to move the DX half up into the top half of EAX, yes.