Monad instance for pairs

问题

This question shows an instance definition for (,) a b, where a is an instance of Monoid.

However, I don't know how to write similar thing for (,) a b, and b is an instance of Monoid? I can basically do this as long as I can write the definition:

instance Monoid b => Monad ((,) ???) where
    return a = (a,mempty)
    ~(a,b) >>= f = let (c,b1) in f a in (c,b `mappend` b1)

So the question is how to write the ??? part?

UPDATE

Actually this question is special case of a more generic problem: is it possible to write instance of type classes that act on some types that is not appear at the end? In my case, the type constructor is (,) a b, and I want to make it an instance of Monad a where a is not the last type parameter.

回答1:

We could write, for now synonyms works inappropriate for this case, so we use newtype:

newtype RevTuple b a = RevTuple { totuple :: (a , b) }

instance Monoid b => Monad (RevTuple b) where
    return a = RevTuple (a,mempty)
    (RevTuple (a,b)) >>= f = 
                 let RevTuple (c,b1) = f a in RevTuple (c,b `mappend` b1)

回答2:

You cannot. It's crucial for Monad to be polymorphic in its parameter, so if a particular monadish thing only makes sense for monoids, it's not a monad.

For example, return :: Monad m => a -> m a. There's no way you can add an extra constraint to the signature of return.

来源：https://stackoverflow.com/questions/19196333/monad-instance-for-pairs