Why MonadPlus and not Monad + Monoid?

Question

I\'m trying to understand the motivation behind the MonadPlus. Why is it necessary if there are already the typeclasses Monad and Monoid

Answer 1

With the QuantifiedConstraints language extension you can express that the Monoid (m a) instance has to be uniform across all choices of a:

{-# LANGUAGE QuantifiedConstraints #-}

class (Monad m, forall a. Monoid (m a)) => MonadPlus m

mzero :: (MonadPlus m) => m a
mzero = mempty

mplus :: (MonadPlus m) => m a -> m a -> m a
mplus = mappend

Alternatively, we can implement the "real" MonadPlus class generically for all such monoid-monads:

{-# LANGUAGE GeneralizedNewtypeDeriving, DerivingStrategies, QuantifiedConstraints #-}
{-# LANGUAGE UndecidableInstances #-}

import Control.Monad
import Control.Applicative

newtype MonoidMonad m a = MonoidMonad{ runMonoidMonad :: m a }
    deriving (Functor, Applicative, Monad)

instance (Applicative m, forall a. Monoid (m a)) => Alternative (MonoidMonad m) where
    empty = MonoidMonad mempty
    (MonoidMonad x) <|> (MonoidMonad y) = MonoidMonad (x <> y)

instance (Monad m, forall a. Monoid (m a)) => MonadPlus (MonoidMonad m)

Note that depending on your choise of m, this may or may not give you the MonadPlus you expect; for example, MonoidMonad [] is really the same as []; but for Maybe, the Monoid instance lifts some underlying semigroup by artifically giving it an identity element, whereas the MonadPlus instance is left-biased choice; and so we have to use MonoidMonad First instead of MonoidMonad Maybe to get the right instance.