Rick Myers wrote: >On Jan 28, 2002 at 19:56:05 -0600, Dave Hoover wrote: >> 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.
push @things_to_learn, 'binary'; Thanks for trying, --Dave
