Modulo

[joke]

ive been reading this artice for fat
http://scottie.20m.com/fat.htm

It says :

Now that we know the logical sector, you might be wondering
how to get it to absolute sector so we can read it. Well
we use a few formulas to figure it out.
They are:
sector = ( logical sector MOD sector per track) + 1
head = (logical sector \ sector per track ) MOD number of heads
track = logical sector \ (sector per track * number of heads)

Note that this is integer division instead of floating point
division. And also Modulo math. If you can't figure this out
just turn on QBasic and enter a quick formula.
So. Using the formulas above we get:
sector= (34 MOD 512)+1= 17
head= (34 \ 512) MOD 2= 1
track= 34 \ (512 *2)= 0
Our file starts at Head 1 Track 0 Sector 17
The next sector Head 1 Track 0 Sector 18
The next sector Head 0 Track 1 Sector 1
and so on till the end.

My question is how can i implement modulo, i mean assembly doesn't have a modulo instruction. And mathematicaly modulo has many meanings. Please tell me is there a way to do modulo in assembly and if not can i do the above calculations without modulo?

Best Regards

[matthias]

I don't know if this will help you but I have written some code for you in c:

Code: Select all

int modulo(int a, int b)
{
	return (a % b);
}

Disassembly:

Code: Select all

00000000 <_modulo>:
   0:	55                   	push   %ebp
   1:	89 e5                	mov    %esp,%ebp
   3:	8b 55 08             	mov    0x8(%ebp),%edx
   6:	89 d0                	mov    %edx,%eax
   8:	c1 fa 1f             	sar    $0x1f,%edx
   b:	f7 7d 0c             	idivl  0xc(%ebp)
   e:	89 d0                	mov    %edx,%eax
  10:	5d                   	pop    %ebp
  11:	c3                   	ret 

[carbonBased]

My question is how can i implement modulo, i mean assembly doesn't have a modulo instruction. And mathematicaly modulo has many meanings. Please tell me is there a way to do modulo in assembly and if not can i do the above calculations without modulo?

Assembly does have a modulo (aka modulus) function. Take a closer look at your intel instruction set docs -- particularly div and idiv.

You're lookin' for dx.

--Jeff

[SpooK]

The modulo operator is a logical abstraction, derived from mathematics, that specifies the remainder of a division operation.

On the x86, this is achieved by the use of the DIV/IDIV instruction(s) with the source number (base) to divide in EAX. Using a 32-bit divisor (i.e. div ebx), EDX:EAX will be divided by EBX, and the result will be put into EAX with the remainder (modulo) in EDX... just make sure EDX is zeroed-out before performing the DIV.

Now, since you mentioned Assembly Language, and you are using powers of two (2 and 512) in your formula... you may wish to use a simple bitwise AND instead of a time consuming division operator.

Code: Select all

;Modulo example for "Sector", power of 2
mov eax, 34 ;Base
mov ebx, 511 (512-1) ;Divisor->Bitmask
and eax,ebx; Remainder in EAX

This code idea is good to keep in mind, but only use it if you are interested in just the remainder. It would take a few more instructions to implement result+remainder, probably not worth the effort

[carbonBased]

This code idea is good to keep in mind, but only use it if you are interested in just the remainder. It would take a few more instructions to implement result+remainder, probably not worth the effort

Actually, it's probably entirely worh the effort, in terms of speed. DIV and IDIV are two of the slowest instructions on an Intel CPU (and most others, I assume).

If working with powers of two, I definitly recommend computing division and modulo using shifts and ands. Just make sure to comment what your intention is, for those that don't see why you're doing it.

--Jeff