为什么MonadFix的实例不能用于连续monad？

问题描述

我们如何证明没有？

How can we prove that the continuation monad has no valid instance of MonadFix?

推荐答案

不是不能有一个 MonadFix 实例，只是库的类型有点过于拘束。如果您在所有可能的 r s中定义 ContT ，则不仅 MonadFix 成为可能，但是所有高达 Monad 的实例都不需要底层仿函数：

Well actually, it's not that there can't be a MonadFix instance, just that the library's type is a bit too constrained. If you define ContT over all possible rs, then not only does MonadFix become possible, but all instances up to Monad require nothing of the underlying functor :

newtype ContT m a = ContT { runContT :: forall r. (a -> m r) -> m r }
instance Functor (ContT m) where
  fmap f (ContT k) = ContT (\kb -> k (kb . f))
instance Monad (ContT m) where
  return a = ContT ($a)
  join (ContT kk) = ContT (\ka -> kk (\(ContT k) -> k ka))
instance MonadFix m => MonadFix (ContT m) where
  mfix f = ContT (\ka -> mfixing (\a -> runContT (f a) ka<&>(,a)))
    where mfixing f = fst <$> mfix (\ ~(_,a) -> f a )

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