Haskell 使用穷人与延续/自由单子的并发？

我想创建一个带有挂起函数的交错线程，我可以在调用时传递单个函数

基本延续类型：

data ContT m a = ContT {runContT :: (a -> m a) -> m a}

suspendedFunction :: Int -> ContT Maybe Int
suspendedFunction i = ContT $ \todo -> todo i

calculation :: Int -> Maybe Int --(a -> m a)
calculation i = Just 100

具有穷人并发性的继续：

data C m a = Atomic (m (C m a)) | Done a

instance Monad m => Monad (C m) where
  (>>=) (Atomic m) f      = Atomic $ (liftM (>>= f) m)
  (>>=) (Done a) f        = f a
  return                  = Done

atomic :: Monad m => m a -> C m a
atomic m = Atomic $ liftM Done m

interleave :: Monad m => C m a -> C m a -> C m a
interleave (Atomic m1) (Atomic m2) = do
  n1 <- atom $ m1
  n2 <- atom $ m2
interleave n1 n2
interleave m1 (Done _)             = m1
interleave (Done _) m2             = m2

runInterleavedThread :: Monad m => C m a -> m a
runInterleavedThread m = m >>= runInterleaved

createSuspendedThread :: Int -> C (ContT Maybe) Int
createSuspendedThread i = atomic $ suspendedFunction i

main = do
 let inter = interleave (createSuspendedThread 1) (createSuspendedThread 2)
 let complete = runContT $ runThreads inter $ calculation

---

我是否缺少一些基本的东西，有人能解释一下吗？

“没有实例（Monad（可能是ContT））

有吗？我在任何地方都没有看到这样的实例。如果不想定义自己的实例，应该使用

 "No instance for (Monad (ContT Maybe))
    arising from a use of ‘atomic’"