Question

鉴于下面的概念代码证明，我希望能够以某种方式执行我的foo函数，并能够输出字符串Paul!以及获取其返回值的可能性InputT之后unsafePerformIO monad-transformer未使用runExceptT删除IO包装器。

import Control.Monad.Except

import System.IO.Unsafe (unsafePerformIO)
import System.Console.Haskeline


type ErrorWithIO = ExceptT String IO


foo :: String -> ErrorWithIO String
foo "paul" = do liftIO $ putStrLn "Paul!"
                return "OK!"
foo _ = throwError "ERROR!"


runRepl :: IO ()
runRepl = runInputT defaultSettings $ loop


loop :: InputT IO ()
loop = do
    line <- getInputLine "> "
    case line of
        Nothing -> return ()
        Just input -> do return $ putStrLn "asd"
                         case unsafePerformIO $ runExceptT $ foo input of
                             Left err -> outputStrLn err >> loop
                             Right res -> do
                                 x <- outputStrLn . show $ res
                                 loop




main :: IO ()
main = runRepl >> putStrLn "Goodbye!"

我错过了一些明显的东西吗？

Answer 1

由于InputT IO是MonadIO，您可以liftIO使用此类型：

liftIO :: IO a -> InputT IO a

所以，

do ...
   x <- liftIO $ runExceptT $ foo input
   case x of
     Left err  -> ...
     Right res -> ...

或者，请改用Control.Monad.Trans.lift。

在InputT monad中的Haskeline中执行简单的IO，而不必求助于unsafePerformIO

1 个答案: