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+newtype Cont a = Cont
+  { unCont :: forall r. (a -> r) -> r
+  }
+
+-- I could not figure out `Functor`. After peeking Sandy's solution I was able
+-- to figure out `Applicative`, but failed miserably with `Monad`. I added
+-- annotations on the types of all the terms on my attempt to figure it out.
+
+-- Exercise 6.4-i
+-- Provide a `Functor` instance for `Cont`.
+instance Functor Cont where
+  -- fmap      :: (a -> b) -> Cont a -> Cont b
+  -- f         :: a -> b
+  -- a         :: (a -> r) -> r
+  -- c         :: b -> r
+  -- c . f     :: a -> r
+  -- a $ c . f :: r
+  fmap f (Cont a) = Cont $ \c -> a $ c . f
+
+-- Exercise 6.4-ii
+-- Provide a `Applicative` instances for `Cont`.
+instance Applicative Cont where
+  pure a = Cont $ \c -> c a
+
+  -- (<*>) :: Cont (a -> b) -> Cont a -> Cont b
+  -- f     :: ((a -> b) -> r) -> r
+  -- a     :: (a -> r) -> r
+  -- c     :: b -> r
+  -- c'    :: a -> b
+  Cont f <*> Cont a = Cont $ \c -> f $ \c' -> a $ c . c'
+
+-- Exercise 6.4-iii
+-- Provide a `Monad` instances for `Cont`.
+instance Monad Cont where
+  -- (>>=)      :: Cont a -> (a -> Cont b) -> Cont b
+  -- a          :: (a -> r) -> r
+  -- b          :: (b -> r) -> r
+  -- f          :: a -> Cont b
+  -- unCont . f :: a -> (b -> r) -> r
+  -- c          :: b -> r
+  -- c'         :: a
+  Cont a >>= f = Cont $ \c -> a $ \c' -> unCont (f c') c