Haskell function from (a -> [b]) -> [a -> b]

Question

I have a function seperateFuncs such that

seperateFuncs :: [a -> b] -> (a -> [b])
seperateFuncs xs = \\x -> map ($ x) xs
         


        

Answer 1

You can generalize seperateFuncs to Applicative (or Monad) pretty cleanly:

seperateFuncs :: (Applicative f) => f (a -> b) -> (a -> f b)
seperateFuncs f x = f <*> pure x

Written in point-free style, you have seperateFuncs = ((. pure) . (<*>)), so you basically want unap . (. extract), giving the following definition if you write it in pointful style:

joinFuncs :: (Unapplicative f) => (a -> f b) -> f (a -> b)
joinFuncs f = unap f (\ g -> f (extract g))

Here I define Unapplictaive as:

class Functor f => Unapplicactive f where
    extract  :: f a -> a
    unap     :: (f a -> f b) -> f (a -> b)

To get the definitions given by leftaroundabout, you could give the following instances:

instance Unapplicative [] where
    extract = head
    unap f = [\a -> f [a] !! i | i <- [0..]]

instance Unapplicative ((->) c) where
    extract f = f undefined
    unap f = \x y -> f (const y) x

I think it's hard to come up with a "useful" function f :: (f a -> f b) -> f (a -> b) for any f that isn't like (->).