modulo in assembly?

hedpe · 11-19-2006, 08:07 PM

Hey guys,

I am programming on the Intel IXP platform in microcode, and need to get the modulo of two numbers but can't quite figure this one out. They have given a divide() macro to simplify a division operation, but nothing for getting a remainder.

For instance... I have a general purpose register with 18 in it... and another register with 8 in it.

I would like to get a result of 2, with 8 bits in a byte, I want to access the second bit in the third byte... however for now I just want to get "2", then i'll worry about bitwise operations to access it.

I'm not looking for anyone to look up the opcodes for IXP, just looking for a logic answer like "you have to shift left by blah and then do blah"

I'd greatly appreciate it.

Thanks!
George

randyding · 11-19-2006, 08:44 PM

If you already know the quotient Q=N/D then the remainder can be calculated R=N-D*Q

paulsm4 · 11-19-2006, 10:05 PM

For X86 assembly, of course, it's really, really easy:

Quote:

DIV: unsigned division

Quote:

IDIV: signed division

Code:

Dividend  Dividend size  Quotient  Remainder
--------  -------------  --------  ---------
AX        16 bits        AL        AH
DX:AX     32-bits        AX        DX
EDX:EAX   64-bits        EAX       EDX

The "modulus" (C/C++ "%" operator) would just be the remainder of integer division (in X86 assembler!).

As far as your microcode (without a real "divide" operation): Randyding's suggestion sounds reasonable.