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我从 得到一个结果getInputline，其类型是：

(MonadException m) => IO String -> InputT m (Maybe String)

我只想得到那Maybe String部分。我很清楚，一般来说，没有办法剥离单子，正如这个答案（以及同一问题中的其他答案）中所解释的那样。但是，由于我是在 内部进行的InputT，所以我想这是可能的，正如这里所建议的那样。但是，正如答案所暗示的那样，我不能只使用liftIO，因为IO在StateT.

loop :: Counter -> InputT (StateT [String] IO) ()    
loop c = do
        minput <- getLineIO $ in_ps1 $ c
        case minput of
          Nothing -> outputStrLn "Goodbye."
          Just input -> (process' c input) >> loop c      

getLineIO :: (MonadException m) => IO String -> InputT m (Maybe String)
getLineIO ios = do
    s <- liftIO ios
    getInputLine s

process' :: Counter -> String -> InputT (StateT [String] IO) ()
[...]

我得到的错误：

Main.hs:81:15:
    No instance for (MonadException (StateT [String] IO))
      arising from a use of ‘getLineIO’
    In the expression: getLineIO
    In a stmt of a 'do' block: minput <- getLineIO $ in_ps1 $ c
    In the expression:
      do { minput <- getLineIO $ in_ps1 $ c;
           case minput of {
             Nothing -> outputStrLn "Goodbye."
             Just input -> (process' c input) >> loop c } }

如果我按照@chepner 的建议直接删除getLineIO并使用：getInputLine

loop :: Counter -> InputT (StateT [String] IO) ()
loop c = do
    minput <- (in_ps1 c) >>= getInputLine
    case minput of
      Nothing -> outputStrLn "Goodbye."
      Just input -> (process' c input) >> loop c

我收到一个错误：

Main.hs:81:16:
    Couldn't match type ‘IO’ with ‘InputT (StateT [String] IO)’
    Expected type: InputT (StateT [String] IO) String
      Actual type: IO String
    In the first argument of ‘(>>=)’, namely ‘(in_ps1 c)’
    In a stmt of a 'do' block: minput <- (in_ps1 c) >>= getInputLine

完整的代码可以在这里找到，关于我想要做什么的解释可以在这里找到。

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