No, that's too easy a cop-out. :-) How would you stop at the right number in Python code? Would you compute 100 numbers and then inspect/truncate afterwards?

Ah, I see that I truly missed your earlier point. I suppose I would just do this:

  def fib(max):
      a, b = 0, 1
      while a < max:
          yield a
          a, b = b, a + b

  for k in fib(100):
      print(k)