Monadic function of `(a -> m (Either e b)) -> Either e a -> m (Either e b)`?

user_11801882 0
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Is there a way to write this function in a more "monadic" way, instead of resorting to pattern matching on Either?

{-# LANGUAGE LambdaCase #-}

calculate :: (Monad m) => (a -> m (Either e b)) -> Either e a -> m (Either e b)
calculate f = \case
  Left err   -> return $ Left err
  Right vals -> f vals

Specifically, for my use case, m is IO; f is a function that takes in input and produces some IO effect or fails, and the input is something that could have failed already.

Maybe using ExceptT?

either monads haskell
2 Answers

Yep, looks like ExceptT to me. Though I would probably not use a function with this signature -- instead, I would use ExceptT more broadly, and then this function is just (=<<). Of course this is guesswork based on the use case.

But if you must:

calculate :: (Monad m) => (a -> m (Either e b)) -> Either e a -> m (Either e b)
calculate f m = runExceptT (ExceptT . f =<< ExceptT m)
user_337960

You can use traverse and join:

calculate :: (Monad f, Traversable f, Applicative m) => (a1 -> m (f a2)) -> f a1 -> m (f a2)
calculate f e = join <$> traverse f e

Note the more general type signature. That's what GHC infers. Instead of Either e any type is sufficient as long as it has instances for Monad and Traversable. Also, m doesn't need to have a Monad, Applicative is enough.

ExceptT works as well (the first one doesn't change the type of the result, the second one goes all in on ExceptT):

calculate :: (Monad m) => (a -> ExceptT e m b) -> Either e a -> m (Either e b)
calculate f e = runExceptT $ ExceptT (pure e) >>= f

calculate2 :: (Monad m) => (a -> ExceptT e m b) -> Either e a -> ExceptT e m b
calculate2 f e = ExceptT (pure e) >>= f

I personally would prefer the former, because I find it easier to grasp without ExceptT.

user_52475020
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