2000-05-01-13:12:47 Jeff Pinyan: > $_&$_-1 > > Do you happen to have the proof for why that is 0 for ONLY powers of 2? :) For $_ & f($_) to equal zero, f($_) must clear the highest set bit in $_. Subtracting 1 from a positive integer only clears the highest set bit if the integer is a power of 2, otherwise it just dinks with lower-order bits and leaves the highest one alone. -Bennett