问题
While trying to familiarize myself with Control.Arrow, I have noticed that the Kleisli newtype would seem to admit a Functor instance, something like:
instance Monad m => Functor (Kleisli m a) where
fmap f (Kleisli k) = Kleisli $ liftM f . k
Is there a reason why this instance isn't provided? Does it exist in some package as an orphan instance?
回答1:
Every arrow can be made into a valid Functor by defining
fmap f a = a >>> arr f
However it's not possible to declare a Functor to be a superclass of Arrow because of their different kinds (Functor needs * -> * while Arrow needs * -> * -> *). So every arrow needs to define the instance separately.
You can wrap any arrow with ArrowMonad, which then gives an Applicative instance (and therefore also a Functor): instance Arrow a => Applicative (ArrowMonad a) where ....
I don't see any particular reason why Kleisli lacks the Functor instance. The most probable seems to be that you don't need it. If you want to use functorial (or applicative or monadic) operations, you do it on the original monad. You only wrap the monad into Kleisli when you need the arrow interface.
回答2:
UPDATE
in Control.Arrow is already defined:
(>>^) :: Arrow a => a b c -> (c -> d) -> a b d
(^<<) :: Arrow a => (c -> d) -> a b c -> a b d
UPDATE 2
If you wish to insert Free Monad into Kleisli - it is impossible, Free has one extra parameter f.
So you need to use Arrow Transformer or create a new Arrow class, like
class Arrow a => ArrowFunctor f a | a -> f where
afmap :: a b (f c)
Package arrows consist some examples, but it not implement Free
来源：https://stackoverflow.com/questions/18925045/why-there-isnt-a-functor-instance-for-kleisli-in-control-arrow