% operator (was Re: count of bits set in a long)

[Risto Lankinen]

Oops!

*JUST* having sent the article claiming that Parity(x) would be equal to
Mod(x,1) , I made a second thought.  It is not so, since the result of the
algorithm I showed will be '1' for any 'x!=0' because no two positive
numbers yield zero when added one another.

Terveisin: Risto Lankinen
-- 
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Nokia Data Systems, Technology Dept *  2                              2   *
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