Question

在IO计算中工作我最终得到case mbValue of … s的阶梯，并且发现我应该使用Maybe monad来简化代码。由于它在IO计算范围内且我需要获得IO个值，因此我使用了MaybeT monad变换器，以便lift IO计算到{{} 1}}。

现在，我一直在考虑在Maybe计算中Maybe之后被values <- mbValue“剥离”的值，但事实证明这太简单了这里。

如下所示，当使用Maybe值作为Maybe a（此处将其传递给a）时，无法输入check：

read

如果我为import Control.Monad.Trans (lift) import Control.Monad.Trans.Maybe (runMaybeT) lol :: IO (Maybe Int) lol = return (Just 3) lal :: IO (Maybe String) lal = return (Just "8") foo :: IO (Maybe Bool) foo = do b <- runMaybeT $ do x <- lift lol y <- lift lal return (x < (read y)) return b ^-- Couldn't match type ‘Maybe String’ with ‘String’ main = foo >>= print添加了一个输入的洞，我发现它需要一个return (x < (read y))，这是有道理的，但是当前的绑定还包括

Bool

即，|| y :: Data.Maybe.Maybe GHC.Base.String || (bound at /private/tmp/test.hs:14:5) || x :: Data.Maybe.Maybe GHC.Types.Int || (bound at /private/tmp/test.hs:13:5)是y。这当然解释了错误，但我感到困惑。我的理解在哪里错了，我该如何解决这个错误？

Answer 1

简而言之：用lift构造函数替换MaybeT。

请注意

newtype MaybeT m a = MaybeT { runMaybeT :: m (Maybe a) }

和

lift :: (MonadTrans t, Monad m) => m a -> t m a

您在

中使用lift

x <- lift lol

属于

类型

lift :: IO (Maybe Int) -> MaybeT IO (Maybe Int)

这就是x再次成为Maybe Int的原因。 lift添加了一个新的MaybeT图层，该图层与您已有的Maybe图片无关。

但是

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

而不是

x <- MaybeT lol

将用于

类型

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

做正确的事。

Answer 2

专门针对MaybeT，lift :: Monad m => m a -> MaybeT m a。由于lol :: IO (Maybe Int)，m为IO且a为Maybe Int，因此为lift lol :: MaybeT IO (Maybe Int)。

IO (Maybe a)只是MaybeT IO a newtype包装中包含的值，因此无需解除它;而是使用MaybeT构造函数，例如在MaybeT lol中。

但这不是人们倾向于使用monad变形金刚的方式。相反，只需使用MaybeT值并根据需要提升：

import Control.Monad
import Control.Monad.Trans (lift)
import Control.Monad.Trans.Maybe (runMaybeT, MaybeT)

lol :: MaybeT IO Int
lol = return 3

lal :: MaybeT IO String
lal = return "8"

foo :: IO (Maybe Bool)
foo = 
  runMaybeT $ do
    x <- lol
    y <- lal

    _ <- lift getLine -- lift (IO String) to MaybeT IO String
    _ <- return 100   -- lift any pure value    
    _ <- mzero        -- use the MonadPlus instance to get a lifted Nothing.

    return (x < (read y)) 

main = foo >>= print

MaybeT计算中的类型

2 个答案: