递归liftIO

Question

I've looked at the some instances of MonadTrans, for MaybeT the implementation looks like this:

instance MonadTrans MaybeT where
    lift = MaybeT . liftM Just

As I understand the instance for MonadIO is used to do a variable number of lifts from the inner most, a IO monad, directly to the outermost. For the MaybeT case it looks like this:

instance (MonadIO m) => MonadIO (MaybeT m) where
    liftIO = lift . liftIO

What I don't understand is how this recursive function escapes the infinite loop. What is the base case?

Answer 1

Perhaps surprisingly, the definition below is not recursive, even if it looks such.

instance (MonadIO m) => MonadIO (MaybeT m) where
    liftIO = lift . liftIO

This is because the liftIO on the left hand side is the liftIO for the MaybeT m monad, while the liftIO on the right hand side is the liftIO for the m monad.

Hence, this simply defines liftIO in one monad in terms of the liftIO for another monad. No recursion here.

This is similar to eg

instance (Show a, Show b) => Show (a,b) where
   show (x,y) = "(" ++ show x ++ ", " ++ show y ++ ")"

Above, we define how to print a pair depending on how to print their components. It looks recursive, but it is not really such.

It could help visualizing this by inserting explicit type arguments, at least mentally:

-- pseudo-code
instance (Show a, Show b) => Show (a,b) where
   show @(a,b) (x,y) = 
      "(" ++ show @a x ++ ", " ++ show @b y ++ ")"

Now show @(a,b) , show @a , and show @b are distinct functions.

Answer 2

Simple equational reasoning and rewriting definitions for some specialization can help you. Base case for MonadIO is IO . MaybeT is monad transformer, so lets combine MaybeT and IO in some simple example.

foo :: MaybeT IO String
foo = liftIO getLine

Now let's rewrite this function definition applying instance implementations from your question step by step.

foo
= liftIO {- for MaybeT -} getLine
= lift (liftIO {- for IO here -} getLine)  -- step 2
= lift (id getLine)
= lift getLine
= MaybeT (liftM Just getLine)

1. getLine has type IO String
2. liftM Just getLine has type IO (Maybe String)
3. MaybeT ma constructor needs value of type m (Maybe a) where m = IO and a = String in our case.

Probably hardest step to analyze is step 2. But in reality it's very easy if you remind yourself types of liftIO :: IO a -> ma and lift :: Monad m => ma -> tma . So all work is done by type inference.