Re: implementing software unsigned divide

[Stephen Leake <stephen.a.leake.1@gsfc.nasa.gov>]

Tucker Taft <stt@averstar.com> writes:

> Does your chip happen to have 32-bits/16-bits => 16-bit
> quotient and 16-bit remainder?  

No.

> For that case, clearly you just zero out the high order part rather
> than sign-extending to handle an unsigned dividend. If it has 32/32
> => 32, of course you can handle 16-bit unsigned trivially.

Yes, I got that part :).

I also realized (after your message yesterday) that if neither operand
has the high bit set, the signed divide will give the correct answer.

> A more general solution, I believe, is to (logical) shift both the
> dividend and divisor right 1 bit, do the divide, and then adjust the
> quotient up or down by 1 to get the remainder to be >= 0 and less
> than the divisor. (Obviously you need to first check for the special
> case when the divisor is <= 1!) 

Ah, that sounds promising. Thanks!


-- 
-- Stephe