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Comment by jparise | Hacker News Reader

Here's a (rough) Python equivalent:

  from itertools import islice

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

  for k in islice(fib(), 1, 100):
      print(k)

It uses a generator function to produce an infinite iterable sequence. islice() is used to produce just a slice of that sequence.

This could probably be written using fewer characters, but I don't think it could be anywhere near a short (and terse) as the Perl version.

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Correct me if I'm wrong, but that code prints 100 fibonacci numbers, whereas the Perl 6 code prints all fibonacci numbers up to, but not including, 100.

You're right. I misread the range of the Perl version. Adjust the slice accordingly.

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)

And a haskell version (not the shortest possible one):

  takeWhile (< 100) fibs
        where fibs = 0 : 1 : zipWith (+) fibs $ tail fibs

I don't think fibs can be expressed as a list comprehension in haskell, can it? It's cute that perl6 is able to use one, but OTOH I can't think of many other lists I'd want to use that trick to calculate.

And, despite having written perl for 15 years, I find the haskell version much clearer, and its use of laziness and infinite lists (zipping them up in particular) much more advanced than perl6's.

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