On Jan 28, 2002 at 19:56:05 -0600, Dave Hoover wrote: > > BooK wrote: > > So we print $_ multiplied by 1 if all of those have > > 1 as their low-order > > bit. And by 0, if any of those numbers is even. > > Can someone explain to my poor ignorant head what "low-order bit" means? > I'm guessing it has something to do with the '&' operator. Be gentle.
Simple. As each requirement for this hole was supposed to be "even", it had to be a multiple of two. Thus, in binary terms, the "low-order-bit" would always be zero. Of course though, if you don't grok binary, that probably doesn't make any sense. > > ~$. is odd if $. is even > > What is this? Tell me more, I'm lost. I didn't get that part myself... --rick
