Question

我从getInputline获得了一个结果，其类型为：

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

我想获得Maybe String部分。我很清楚，正如this answer（以及同一问题中的其他答案）中所解释的那样，一般情况下无法剥离monad。但是，由于我在InputT内进行此操作，我认为这是可能的，正如提议here一样。但是，正如答案所示，我无法使用liftIO，因为IO位于StateT内。

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 } }

如果我删除getLineIO并直接使用getInputLine，请按照@Chepner的建议：

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

可以找到完整的代码here，并且可以找到关于我尝试做什么的解释here。

从输入中解包getInputLine结果

0 个答案: