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Thread: Division routines.
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face BY : Bob Ammerman email (remove spam text)



I am pretty sure something like the following will work...

Let's say you have a number of bits 'N', typically 8 or 16.

Then, many division routines will have a dividend of 2N bits, and a divisor
of N bits.

So, say you want to divide a N-bit number by another N-bit number with an
implied binary point on the left...

Simply extend your N-bit dividend by N zeros on the right and do the
division.

~ Bob Ammerman
RAm Systems

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