Re: 'Divide by A' procedure in 90x0 ROM

From: Michał Pleban <>
Date: Sat, 19 Feb 2011 19:18:04 +0100
Message-ID: <>

W dniu 2011-02-19 01:31, Rhialto pisze:

> I don't actually think that Accumulator+1 is really needed, unless the
> dividend would be divided by values over 0xFF. After all, the
> Accumulator is where the high bits of Result are shifted into, and when
> the divisor can be subtracted from it, it will be. So this value will
> never be bigger than the divisor, which (here) is 1 byte.

It is needed because the accumulator must always be 1 bit wider than the
divisor. So you need 9 bits for it in this case (imagine dividing $FE00
by $FF).


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Received on 2011-02-19 19:00:06

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