[Haskell-beginners] list monad question

[Daniel Fischer]

Am Freitag 30 Oktober 2009 14:32:35 schrieb Matthias Guedemann:
> Hi Daniel,
>
> > That gives
> >
> > combinations n xs = foldr f [[]] (replicate n xs)
> >
> > pointfree, for extra goodness:
> >
> > -- pointfree f inline
> > combinations n xs = foldr ((. (. (:)) . flip map) . (>>=)) [[]]
> > (replicate n xs) -- eliminate xs
> > combinations n = foldr ((. (. (:)) . flip map) . (>>=)) [[]] . replicate
> > n -- completely pointfree
> > combinations = (foldr ((. (. (:)) . flip map) . (>>=)) [[]]  .) .
> > replicate
>
> thank you, looks rather strange to me but works well.

Yes :D The pointfree f is nicely obfuscated. But if your friend is a perl coder, he should 
be able to appreciate that.
The standard way to express f, however, is liftM2 (:), so

combinations = (foldr (liftM2 (:)) [[]] .) . replicate 
-- isn't that boring?

But earnestly, replicateM is the thing to use.

>
> regards