20个中级Haskell练习-练习9

Question

I was working on 20 intermediate haskell exercise. And Exercise 9 is:

class Misty m where
    banana :: (a -> m b) -> m a -> m b
    unicorn :: a -> m a

And the solution offered here is

instance Misty ((->) t) where
    banana x y z = (x $ y z) z
    unicorn f _ = f

If I replace m with ((->) t) in banana's type signature, I get

(a -> (t->b)) -> (t->a) -> (t->b)

It seems a binding function for ((->) t). So why does banana needs three arguments? And why the answer is (x $ yz) z?

Thank you for help. :)

Answer 1

The key insight is that the signature

(a -> (t->b)) -> (t->a) -> (t->b)

Is equivalent to:

(a -> (t->b)) -> (t->a) -> t -> b

This means that the binding banana xyz binds the following:

x :: a -> (t->b)
y :: t -> a
z :: t

And therefore:

y z :: a
x $ y z :: t -> b
(x $ y z) z :: b