Need some quick 68K assembly help...

[slugg]

I can't seem to find much about 68K asm, and I'm not sure if what I have is correct... But lets say we have 2 positive integers of word length in the first two data registers... like so:

MOVE.W #10,D0
MOVE.W #3,D1

So now I wanna find the remainder of dividing D0 by D1. I can't find any instruction for modulus, so I figured I need to get creative... My idea is as follows:

edit: fixed error pointed out by ChristianV

CLR.L D2
MOVE.W D0,D2
DIVU D1,D2
SWAP D2


^^ so basically:
- Clear out all 32 bits of D2
- Copy the contents of D0 into D2's last 16 bits (word length)
- Divide D2 by D1... D2 should now have the remainder in the upper 16 bits? (i think)
- Swap the upper and lower words of D2...
- And now I can refer to D2.W as the remainder of D0 divided by D1?

^^ This seems okay in theory... but I really don't know. Any advice?

edit: This is confirmed working. Thanks. I'll leave this hear in the off chance that someone's searching for the same solution.

 

[DaveSimmons]

It's been years since I've done any assembly, and it was 6502, 8080 and x86. But for intel I think some division operations store the remainder into another register, and remainder == modulus. Check whether 68k division does this too.

Brute force would be something like:
do ( if a > b, then a -= b ) while (a > b)

 

[ChristianV]

Yes, on Intel it works like this, but in 68k assembly, the remainder is stored in bits 16-31, the result in bit 0-15, in the same register. The SWAP command is the right one to use. But isn't the result stored in D1? because, you can execute DIVU #4,D1 (I assume #4 means that it is an immediate), so the result is stored in D1, obviously. That's why I think it has to be D1, too, in your example. However, I'm not sure because I've never written anything in 68k assembly.

 

[slugg]

er yea you're right Christian. Thanks! My code is working