MaybeT/Maybe 和 IO：信息的故障安全读取

Question

我正在尝试读取用户输入的信息并将其解析为Person 类型，该类型使用Gender 类型。为此，我使用以下代码：

data Person = Person String Int Gender String
data Gender = Male | Female | NotSpecified deriving Read

instance Show Gender where
    show Male = "male"
    show Female = "female"
    show NotSpecified = "not specified"

instance Show Person where
    show (Person n a g j) = "Person {name: " ++ n ++ ", age: " ++ show a ++ 
        ", gender: " ++ show g ++ ", job: " ++ j ++ "}"

readPersonMaybeT :: MaybeT IO ()
readPersonMaybeT = do
    putStrLn "Name?:"
    name <- getLine
    putStrLn "Age?:"
    ageStr <- getLine
    putStrLn "Gender?:"
    genderStr <- getLine
    putStrLn "Job?:"
    job <- getLine

    let newPerson = Person name (read ageStr) (read genderStr) job
    putStrLn $ show newPerson

现在我想让这更加安全 - 为了实现这一点，我尝试使用 MaybeT monad。使用这个，我得到了这个代码：

readPersonMaybeT :: MaybeT IO ()
readPersonMaybeT = do
    lift $ putStrLn "Name?:"
    name <- lift getLine
    lift $ putStrLn "Age?:"
    ageStr <- lift getLine
    lift $ putStrLn "Gender?:"
    genderStr <- lift getLine
    lift $ putStrLn "Job?:"
    job <- lift getLine

    let newPerson = Person name (read ageStr) (read genderStr) job
    lift $ putStrLn "show newPerson"

它由 GHCI 编译/加载，但是当我尝试执行 readPersonMaybeT 函数时，我收到错误消息

（Data.Functor.Classes.Show1 IO）没有实例 因使用“打印”而产生 在交互式 GHCi 命令的 stmt 中：打印它

我该如何解决这个问题？编写这段代码时，我使用了关于 Monad Transformers 的wikibook。

编辑：当我尝试使用runMaybeT '运行'它时，它会被执行，但它根本不是故障安全的。例如，输入年龄的废话仍然会导致输出类似于

Person {name: 85, age: *** Exception: Prelude.read: no parse.

Answer 1

如果您仅在要求所有输入后才进行验证，我将只使用 IO monad 并返回一个 Maybe：

import Text.Read
import Control.Monad.Trans.Maybe
import Control.Monad.IO.Class

askPerson :: IO (Maybe Person)
askPerson = do
  name <- putStr "Name? " >> getLine
  a <- putStr "Age? " >> getLine
  g <- putStr "Gender? " >> getLine
  return $ do age <- readMaybe a
              gender <- readMaybe g
              return $ Person name age gender

注意我们如何在 return 语句中使用 Maybe monad。

如果您想在输入无效值后退出询问输入，我会使用 MaybeT --

askPersonT :: MaybeT IO Person
askPersonT = do
  name   <- liftIO $ putStr "Name? " >> getLine
  age    <- MaybeT $ fmap readMaybe $ putStr "Age? " >> getLine
  gender <- MaybeT $ fmap readMaybe $ putStr "Gender? " >> getLine
  return $ Person name age gender

doit = runMaybeT askPersonT

如果用户输入的年龄无效，则不会要求他们输入性别。