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

"Is there some datatype f which has a function :: (a -> f b) -> f (a -> b)?"

In fact, there is an even more general version of this function in the Traversable type class, which deals with commutable functors:

class (Functor t, Foldable t) => Traversable t where

  ... 

  sequenceA :: Applicative f => t (f b) -> f (t b)

How is this related to your function? Starting from your type, with one type substitution, we recover sequenceA:

1. (a -> f b) -> f (a -> b) ==> let t = (->) a
2. t (f b) -> f (t b)

However, this type has the constraint that t must be a Traversable -- and there isn't a Traversable instance for (->) a, which means that this operation can't be done in general with functions. Although note that the "other direction" -- f (a -> b) -> (a -> f b) works fine for all functions and all Applicatives f.