Question

Sometimes I have two functions of the form:

f :: a -> (b1,b2)
h :: b1 -> b2 -> c

and I need the composition g. I solve this by changing h to h’:

h' :: (b1,b2) -> c

Can you please show me (if possible) a function m, so that:

(h . m . f) == (h' . f)

Or another way to deal with such situations. Thanks.

Answer 1

What you’re looking to do is to take a function that operates on curried arguments, h, and apply it to the result of f, which is a tuple. This process, turning a function of two arguments into a function that takes one argument that is a tuple, is called uncurrying. We have, from Data.Tuple:

curry :: ((a, b) -> c) -> a -> b -> c 
   -- curry converts an uncurried function to a curried function.

uncurry :: (a -> b -> c) -> (a, b) -> c
   -- uncurry converts a curried function to a function on pairs.

So now we can write:

f :: a -> (b,c)
f = undefined

h :: b -> c -> d
h = undefined

k :: a -> d
k = uncurry h . f

---

Another tricky way to think of this is via an applicative functor,

k = (h <$> fst <*> snd) . f

Idea from Conor McBride, who’d write it as: (|f fst snd|) . f I think.