On Friday 25 June 2004 10:47, Michael Sokolov wrote:
> Ahmm, but the reciprocal of an integer is not an integer, so you are
> stepping out of integer-land into floating point territory. Do you
> really want to use floating point to implement integer division
> (dividing an integer by an integer to get quotient and remainder)?
> And how do you guarantee that the final integer result will be exact?
> (I guess you convert the dividend to a float, get the reciprocal of
> the divisor, which will be a float, do float multiplication, and
> integerise the result, right? Are there enough bits of precision in
> a float to guarantee an exact result for integer division done this
> way? I doubt that, since AFAIK Alpha has usual 32-bit and 64-bit
> floats, which have other things besides mantissa in those bits, but
> has 64-bit integers.) And how do you get the remainder? Floating
> division (whether direct or via multiplication by the reciprocal)
> doesn't produce a remainder at all.
Well, if it's really an integer, you could use a fixed-point number for
the division (assume that the highest bit is the 2^(-1)bit or 2^0 if
you can get an extra bit). For this to work, you need to have access
to double the number of input bits to the multiplier for its output. I
remember seeing this before...either in my Sr. level architecture class
or the description of how my SBC-16/40 does division, neither of which
sounds right. I need to go back and figure out where I learned this
from.
Remainder is simply quotient - divisor*result, or drop the higher half
of the result and re-do the multiplication (and the resulting higher
half is the remainder).
Pat
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Received on Fri Jun 25 2004 - 11:13:51 BST