Partially Applied Types in Haskell

Question

Based on this question, in this code

data Promise a b =  Pending (a -> b) | Resolved b | Broken

instance Functor (Promise x) where
    fmap f (Pending g)         


        

Answer 1

You are only missing the equations for Resolved and Broken. The only reasonable implementation I can think of is

fmap f (Resolved x) = Resolved (f x)
fmap _ Broken = Broken

Other than that, your code is fine.