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?

Granted, instances of Monoid are concrete types, whereas instances of Monad require a single type parameter. (See Monoid vs MonadPlus for a helpful explanation.) But couldn't you rewrite any type constraint of

(MonadPlus m) => ...

as a combination of Monad and Monoid?

(Monad m, Monoid (m a)) => ...

Take the guard function from Control.Monad, for example. Its implementation is:

guard :: (MonadPlus m) => Bool -> m ()
guard True = return ()
guard False = mzero

I was able to implement it using only Monad and Monoid:

guard' :: (Monad m, Monoid (m ())) => Bool -> m ()
guard' True = return ()
guard' False = mempty

Could someone please clarify the real difference between MonadPlus and Monad + Monoid?

Answer 1

But couldn't you rewrite any type constraint of

(MonadPlus m) => ...

as a combination of Monad and Monoid?

No. In the top answer to the question you link, there is already a good explanation about the laws of MonadPlus vs. Monoid. But there are differences even if we ignore the typeclass laws.

Monoid (m a) => ... means that m a has to be a monoid for one particular a chosen by the caller, but MonadPlus m means that m a has to be a monoid for all a. So MonadPlus a is more flexible, and this flexibility is helpful in four situations:

1. If we don't want to tell the caller what a we intend to use.
   MonadPlus m => ... instead of Monoid (m SecretType) => ...

2. If we want to use multiple different a.
   MonadPlus m => ... instead of (Monoid (m Type1), Monoid (m Type2), ...) => ...

3. If we want to use infinitely many different a.
   MonadPlus m => ... instead of not possible.

4. If we don't know what a we need. MonadPlus m => ... instead of not possible.

Answer 2

Your guard' does not match your Monoid m a type.

If you mean Monoid (m a), then you need to define what mempty is for m (). Once you've done that, you've defined a MonadPlus.

In other words, MonadPlus defines two opeartions: mzero and mplus satisfying two rules: mzero is neutral with respect to mplus, and mplus is associative. This satisfies the definition of a Monoid so that mzero is mempty and mplus is mappend.

The difference is that MonadPlus m is a monoid m a for any a, but Monoid m defines a monoid only for m. Your guard' works because you only needed m to be a Monoid only for (). But MonadPlus is stronger, it claims m a to be a monoid for any a.

Answer 3

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 choice 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.