Modern Electronics (was Re: List charter mods & headcount... ;

From: Fred Cisin <cisin_at_xenosoft.com>
Date: Fri Jun 25 15:05:23 2004

> > Then there are special cases: the Cray and Alpha integer units, which
> > don't offer divide at all. Instead, you're supposed to multiply by
> > the reciprocal of the divisor.
On Fri, 25 Jun 2004, 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

"Most important thing to know about doing floating point arithmetic:
DON'T."

> 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.

REMAINDER = NUMERATOR - ((int)QUOTIENT * DENOMINATOR) ;
which adds an otherwise unnecessary integer multiply and integer
subtraction.
Received on Fri Jun 25 2004 - 15:05:23 BST