haskell - 如何处理一个 IO (Maybe (IO (Maybe t))) 类型？

Question

我正在处理一个库( ghcjs-dom )，其中每个函数都返回一个 IO (Maybe T) .

我有一个函数a with 返回 IO (Maybe x)和功能 b需要x作为参数并返回 IO (Maybe y) .

是一个让我做a ??? b的运算符(operator)并获得 IO (Maybe y) .我的 Hoogle 搜索一无所获。

我看起来像 join适用于 IO (Maybe (IO (Maybe t)))而不是 IO (IO t)或 Maybe (Maybe t) .

Answer 1

据我了解，您有:

a :: IO (Maybe X)
b :: X -> IO (Maybe Y)

IO (Maybe a)之间有密切关系和 MaybeT IO a ，即 MaybeT将一个转换为另一个:

MaybeT :: IO (Maybe a) -> MaybeT IO a

而逆运算只是 runMaybeT :

runMaybeT :: MaybeT IO a -> IO (MaybeT a)

在 MaybeT monad 中，您想要执行的组合只是
绑定(bind)操作:

MaybeT a >>= (\x -> MaybeT (b x)) :: MaybeT IO Y

这会产生 MaybeT IO Y 类型的值。 .将其转换回 IO (Maybe Y)只需使用 runMaybeT .

更新

这是一个“组合” a 和 b 的运算符:

andThen :: IO (Maybe a) -> (a -> IO (Maybe b)) -> IO (Maybe b)
andThen a b = runMaybeT $  MaybeT a >>= (\x ->  MaybeT (b x) )

但是，如果您发现自己经常使用此运算符，也许您
应该重做你的函数，以便你主要在 MaybeT IO 中工作
monad，然后你可以使用 >>=与单个 runMaybeT在外面。