> > > The above snippet is taking advantage of a similar property of binary > > numbers, which are base 2. What the above snippet is doing is checking > to > > see if that last digit is a 0 or not (asking "not n&1" is equivalent to > > asking "n&0", since that digit can only be a 0 or 1). > > Not quite. The &1 is simply checking the least significant bit in the > word. n&0 will always give 0. not n&1 checks whether the last digit > -is- or -isn't- 1. > > S.
Not at all, you mean. I was way off with that statement; I'll plead not guilty by virtue of not having had enough coffee today. <8-( (Wearing the dunce cap. . .) Tony R.
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