Understanding Haskell Type Signature

Question

I have to convert a Haskell type signature into a term. The type signature is:

f :: (a -> b -> c) -> (d -> b) -> (d -> a) -> d -> c

The correct resulting term is:

f g h j x = g (j x) (h x)

and here lies my problem as I understand it g is a function which returns a function which returns c and c is function which returns a function d which returns b and b is a function which returns itself which then returns itself again which then returns c .

Correct me if i am wrong.

What I don't get is why is g taking (jx) as first argument and (hx) as second argument. Shouldn't it be the other way around? Haskell is right associative and h is the secound parameter given to the function f and not j .

Answer 1

g:: a -> b -> c , h:: d -> b , j:: d -> a , and x:: d are all independent arguments to f ; their order implies nothing about how we might end up using them in the definition of f .

To start, we know that f uses its arguments to return a value of type c . But none of the arguments have a value of type c ; the only way to get a value of type c is to use g . But in order to use g , you need arguments of type a and type b , and none of f 's arguments have those types. But we could use h and j to get them, if we had an argument of type d to apply them to, and lo and behold, we do have a value of type d : the argument x ..

f g h j x = let aValue = j x
                bValue = h x
                cValue = g aValue bValue
            in cValue

which can be flattened to the original answer of

f g h j x = g (j x) (h x)

---

If you want to think of the return value of f as being d -> c , rather than just c , you can eliminate x from the definition with some point-free trickery.

f g h j = g <$> j <*> h  -- liftA2 g j h

You can even go a little further to remove h and j as arguments, but the result, though simple, is even more incomprehensible:

f = flip . liftA2

Moral of the story: sometimes point-free style abstracts away distracting details, other times it completely obscures the meaning of the function.