How do you do Multiplication?

[Electroman]

I'm using 68k and I can't figure out how you do multiplication, at the moment I'm just using left shifts and addition but there must be another way? Also if I needed to do none integer multiplication I don't have a clue how I'd go about that .

[Electroman]

Sorted it:
MULS.L D1,D2

[Electroman]

Sorry no its not sorted, I get an Opperand 1 error when i use that code .

[Halsafar]

I hear assemble math in some cases is alot faster than regular.
It would be great to have a Vector class built with operators working in ASM.

[Maven]

I'm using 68k and I can't figure out how you do multiplication, at the moment I'm just using left shifts and addition but there must be another way? Also if I needed to do none integer multiplication I don't have a clue how I'd go about that .

Well in x86 you would use the FPU to do non-int math.
http://www.ray.masmcode.com/tutorial/index.html

mul for unsigned and imul for signed.


To be honest though, shl and shr is the faster way of doing multiplication and division.

[Darkwraith]

However, shl and shr only works if you are multiplying by a number that is base 2

[Electroman]

However, shl and shr only works if you are multiplying by a number that is base 2

Yea but as I said the idea was to use Binary shifts to get as close as I could then do a few additions to get to the exact number .

e.g.

a *9

Shift 'a' 3 times to the left
Then add 'a' onto that and you have it .

[chemicalNova]

So similar to:

LOCAL a:DWORD
LOCAL b:DWORD

push 9
pop a
mov b,a
shl a,3
add b,a

?

chem

[Electroman]

So similar to:

LOCAL a:DWORD
LOCAL b:DWORD

push 9
pop a
mov b,a
shl a,3
add b,a

?

chem

Yea apart from there would be code to control how many shifts would be required and even possibly code so that when adding on numbers it could reuse numbers passed on the shifts to make it even more efficient. But That was just my idea of making it with just using limited tools, I was more searching for a better way to do it if it exists in a single statement but it seemed not .

[chemicalNova]

There could possibly be a 1 line statement which performs this, just not in the 8086 / family of instruction sets.

Try searching for different instruction sets/syntaxes, and reading up on their documentation.

chem

[wossname]

For what its worth I solved this very problem because its a question in my ASM book.

It works on the premise of shifting the multiplicand left by x bits every time you find a 1 in the binary of the multiplier then you just add up all the shifted versions of the multiplicand.

I dare say its farfrom the most efficient longhand method, but it works and I'm proud of it .

;SYNTAX:
;	mov eax, 1357
;	mov ebx, 371
;	CALL Multiply


;-------------------
Multiply PROC USES ecx edi esi
;	Multiplies EAX by EBX and leaves the answer in EAX;
;-------------------

	mov esi, 0				; running total
	mov ecx, 0				; the loop counter and bit test value
	L1:
		bt ebx, ecx			; test the bit
		jnc L2				; jump if the bit is zero
		mov edi, eax			; take a copy of the original eax value
		shl edi, cl			; shift left
		add esi, edi			; add it to the running total
	L2:
		inc ecx
		cmp ecx, 32			; stop after 32 bits
	jb L1
	
	mov eax, esi

	ret
Multiply ENDP

I'm currently grappling with the bizarre way that MUL and DIV work.