William Westfield email (remove spam text)
On Sep 5, 2017, at 10:33 AM, David C Brown <gmail.com> wrote: dcb.home
> in my case the divisor is always greater than the dividend so I
> want a result that is conceptually between 0 and nearly 1 with an implied
> binary point to the left.
It’s equivalent to multiplying your dividend by 256 or 65536 and using 16/32bit division, right? Except that you know you’ll run out of 1 bits comparatively early, and might be able to optimize away the extra registers. Hmm. An interesting problem!
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