Perl 6 in 2010 - a retrospection

http://perlgeek.de/blog-en/perl-6/perl-6-in-2010.html

by perlgeek • 16 years ago

55 28 16 years ago

perlgeek.de

Luyt• 16 years ago

I find the 'Perl6 Periodic Table of the Operators' an intriguing poster. http://www.ozonehouse.com/mark/periodic/

colomon• 16 years ago

Unfortunately it's about two years behind the language now. Otherwise I'd have one hanging on my office wall...

bokchoi• 16 years ago

   # Lazy list of Fibonacci numbers up to (but excluding) 100:
   my @fib := 1, 2, *+* ...^ * >= 100;

bkudria2• 16 years ago

Perl, proudly wearing the "Looks Like Line Noise" hat into 2011!

patrickas2• 16 years ago

I think a more apt retort is "looks like Chinese" because when/if you learn Perl6 I assure you that line will seem natural, readable and meaningful.

Still you are perfectly free to not learn Perl6 nor Chinese :-)

btucker• 16 years ago

Would someone be willing to break this down and explain it?

masak5• 16 years ago

By all means.

    my @fib := ...;

Bind something to an array variable. Assignment tends to follow infinite lists into the abyss. (But would also work, in this case, since the list is lazily evaluated.)

    1, 2, *+* ...^ * >= 100;

This is all governed by the `...^` in the middle, which is a lazy list generator. Three components, the seed, the iteration and the end condition, surround it. (Those three are much like a C `for` loop, come to think of it.)

The seed is simply `1, 2`.

The iteration is star-plus-star (which I can't figure out how to write out explicitly in this markup). This is ultimately syntactic sugar for a lambda with two parameters: `-> $a, $b { $a + $b }`. Or, put differently, this is a curried addition function.

The end condition is `* >= 100`. Also curried; we could desugar it as either `-> $c { $c >= 100 }` or simply `{ $_ >= 100}`. It's a block that takes one argument and checks whether it's 100 or greater.

The sequence operator `...^` has a little hat on it, meaning we're to discard the item that eventually matches the end condition. This is congruent with the fact that there is a range operator `..^` (note: two dots) which excludes its right endpoint.

jparise3• 16 years ago

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.

masak5• 16 years ago

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.

joeyh• 16 years ago

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.

perlgeek• 16 years ago

reddit and hackernews are really similar: both pick a single line or two out of a >1k words post and ignore the rest. Impressive.

Perl 6 in 2010 - a retrospection

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