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Re: solutions

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Dave Hoover
January 28, 2002 17:59
Re: solutions
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BooK wrote:
> So we print $_ multiplied by 1 if all of those have
> 1 as their low-order
> bit. And by 0, if any of those numbers is even.

Can someone explain to my poor ignorant head what "low-order bit" means?
I'm guessing it has something to do with the '&' operator.  Be gentle.

> ~$. is odd if $. is even

What is this?  Tell me more, I'm lost.



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