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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
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