Functors Compose, Monads Do Not

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Functor Composition

Let’s start with a refresher of map:

map :: forall a b. (a -> b) -> f a -> f b

In other words, map takes a function a -> b and gives us a function f a -> f b. For that reason, we can take any two nested functors (e.g. Array and Maybe) and run a function on the nested values by putting two maps together:

v :: Array (Maybe Int)
v = [Just 1, Nothing, Just 3]

f1 :: Int -> String
f1 = show

main :: Effect Unit
main = do
 logShow $ map (map f1) v

-- [(Just "1"),Nothing,(Just "3")]

Monad Composition

This time we want to take a look at bind:

bind :: forall a b. m a -> (a -> m b) -> m b

If we tried to compose the same way we did with functors, we would notice the code does not compile:

v :: Array (Maybe Int)
v = [Just 1, Nothing, Just 3]

f2 :: Int -> Array (Maybe String)
f2 i = [Just $ show i]

main :: Effect Unit
main = do
 logShow $ bind v (\x -> bind x f2) -- DOES NOT COMPILE!!

The problem here is in the nested bind:

bind v (\x -> bind x f2)
     ^ Maybe Int
       ^ Int -> Array (Maybe String)

In fact, Maybe Int -> (Int -> Array (Maybe String)) -> ?? is not what bind expects: the first argument seems to indicate that m is Maybe but the second seems to indicate that m is Array. This does not compile since the monad m is supposed to be the same.

To make the program compile we have to make use of a function (i.e. maybe) specific to the monad we are dealing with (i.e. Maybe):

main :: Effect Unit
main = do
 -- logShow $ bind v (\x -> bind x f2) -- DOES NOT COMPILE!!
 logShow $ bind v (maybe (pure Nothing) f2)

Or we could use the MaybeT monad transformer:

v2 :: MaybeT Array Int
v2 = MaybeT [Just 1, Nothing, Just 3]

f3 :: Int -> MaybeT Array String
f3 i = MaybeT [Just $ show i]

main :: Effect Unit
main = do
 --logShow $ bind v (\x -> bind x f2)
 logShow $ bind v (maybe [Nothing] f2)
 logShow $ runMaybeT $ bind v2 f3

Outro

I’ve blatantly copied the content of this blog post out of a talk by Tony Morris. So be sure to check the original stuff out!

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