很明显，Concurrent a与Cont Action a相同，其中Cont是延续单子。下面是对延续的一个简单解释：
1.考虑函数f :: a -> b，我们想把这个函数转换成延续传递风格，我们该怎么做呢？
1.假设我们有一个延续k :: b -> r，它把f的返回值作为输入，并且它本身返回一个任意类型的值r，然后我们可以把f转换成CPS。
1.令g :: a -> (b -> r) -> r是f的CPS版本函数。注意，它接受一个额外的参数（即，延续k），并返回k应用于其输出b的结果。
让我们举一个实际的例子，其中f是 predicate 函数odd :: Int -> Bool：

odd :: Int -> Bool
odd n = n `mod` 2 == 1

下面是以延续传递方式编写的同一个函数：

odd' :: Int -> (Bool -> r) -> r
odd' n k = k (n `mod` 2 == 1)

(Bool -> r) -> r部分可以抽象为延续单子：

data Cont r a = Cont { runCont :: (a -> r) -> r }

odd' :: Int -> Cont r Bool
odd' n = return (n `mod` 2 == 1)

instance Monad (Cont r) where
    return a = Cont (\k -> k a)
    m  >>= f = Cont (\k -> runCont m (\a -> runCont (f a) k))

注意，对于任意类型的r，延续k的类型是Bool -> r。因此，延续k可以是任何以Bool为参数的函数。例如：

cont :: Bool -> IO ()
cont = print

main :: IO ()
main = odd' 21 cont

然而，在Concurrent的情况下，这个r不是任意的。它已经被专门化为Action。实际上，我们可以将Concurrent定义为Cont Action的类型同义词，如下所示：

type Concurrent = Cont Action

现在，我们不需要实现Concurrent的Monad示例，因为它与上面定义的Cont r的Monad示例相同。

runConcurrent :: Concurrent a -> ContinuationPseudoMonad a
runConcurrent (Concurrent g) = g

instance Monad Concurrent where
    return a = Concurrent (\k -> k a)
    m  >>= f = Concurrent (\k -> runConcurrent m (\a -> runConcurrent (f a) k))

注意，在instance Monad Concurrent的定义中，我们没有使用Action，这是因为Concurrent = Cont Action和Cont r的monad示例透明地使用r。

You seem to be reaching for some vocabulary, which is a hard question to phrase. Let's break down what you have into steps, and see if that helps.

data Action = Atom (IO Action)
            | Fork Action Action
            | Stop

Action is an algebraic data type with three constructors. It is a corecursive data type as it is defined in terms of itself.

type Continuation a = a -> Action

Continuation a is a type alias for the function typea -> Action . It is an example of a contravariant functor , since we could define a function

contramap :: (a -> b) -> Continuation b -> Continuation a
contramap aToB bToAction = aToAction 
  where aToAction = \a -> bToAction (aToB a)

Note the reversal - contramap takes a function a -> b and creates a function Continuation b -> Continuation a .

type ContinuationPseudoMonad a = Continuation a -> Action

ContinuationPseudoMonad a is another type alias for a function type, but since Continuation a is also a function type, ContinuationPseudoMonad a is a type of higher-order function, since it takes a function as an argument.
ContinuationPseudoMonad a is also a functor, but it's a covariant functor, as we could define a function

fmap :: (a -> b) -> ContinuationPseudoMonad a -> ContinuationPseudoMonad b
fmap aToB aToActionToAction = bToActionToAction
  where bToActionToAction = \bToAction -> aToActionToAction (\a -> bToAction (aToB a))