将Haskell函数转换为无点表示

Question

I want to create a pointfree function that takes a list of functions, applies a single argument to each listed function, and then compresses the list via another function. A pointfree version of this function would have the following type signature:

multiplex :: ([a] -> b) -> [(c -> a)] -> (c -> b)

And an example usage:

invariantsHold :: (Integral a) => a -> Bool
invariantsHold = multiplex (all id) [(>=0),(<=50),even]

I was able to write the following:

multiplex :: ([a] -> b) -> [(c -> a)] -> c -> b
multiplex f xs e = f $ map ((flip ($)) e) xs

This implementation is not pointfree, how can I transform this function to a pointfree representation?

Answer 1

Not in pointfree style, but surely could be simplified significantly by using Applicative (need to import Control.Applicative ):

multiplex f xs e = f $ xs <*> pure e

invariantsHold also could be simplified a little:

invariantsHold = multiplex and [(>=0),(<=50),even]

---

Using sequenceA (from Data.Traversable ) is another way to define this multiplex :

multiplex f xs e = f $ sequenceA xs e

And this definition can be rewritten in pointfree style (give by pointfree ):

multiplex = (. sequenceA) . (.)

or

multiplex = flip ((.) . (.)) sequenceA

Beautiful, but seems pointless to me :-)