The type of the last part of the expression is:

many $ Just 1 :: Num a => Maybe [a]

So in order to be able to return the “Just” constructor which inspected by the application of (take 3 <$>) we have somehow to know for sure that all the <*> executions will indeed see a “Just” in both of their arguments. This forces more and more evaluations.
  
 Doaitse


Op 29 sep. 2016, om 22:28  heeft Jake <jake.waksbaum@gmail.com> het volgende geschreven:

take 3 $ many $ Just 1

doesn't type check. Did you mean this?

take 3 <$> (many $ Just 1)
I think this may have something to do with the default definition of many in the definition of Alternative:
many :: f a -> f [a]
many v = many_v
  where
    many_v = some_v <|> pure []
    some_v = (fmap (:) v) <*> many_v
many_v and some_v are mutually recursive functions, and it may be that this prevents the thunks from being made available to take in some way. I'm really not sure though, this is just an idea about why this is not quite the same as (take $ repeat 1)

On Thu, Sep 29, 2016 at 3:51 PM Corentin Dupont <corentin.dupont@gmail.com> wrote:
Hi guys,
I'm playing with the mysterious "some" and "many" from Control.Applicative.
If I try:

many $ Just 1

It just loops, I understand why:
http://stackoverflow.com/questions/18108608/what-are-alternatives-some-and-many-useful-for
It seems that some and many are usually used in a context where something is consumed, and can be depleted, so the loop ends.

But why doesn't this terminates?

take 3 $ many $ Just 1

It's a recursive call, but the construction of the result should be lazy...

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