Question

我正在尝试读取用户输入的信息并将其解析为使用类型Person的{{1}}类型。为此，我使用此代码：

Gender

现在我想让这更安全 - 为了达到这个目的，我尝试使用MaybeT monad。使用这个，我得到了这段代码：

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

它由GHCI编译/加载，但当我尝试执行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"函数时，我收到错误消息

没有（Data.Functor.Classes.Show1 IO）的实例使用'print＆＃39; 在交互式GHCi命令的stmt中：打印它

我该如何解决这个问题？编写此代码时，我使用wikibook关于Monad Transformers。

编辑：当我尝试'＃39;它与readPersonMaybeT一起执行，但它根本不是故障安全的。例如，为年龄输入废话仍然会产生类似

的输出

人{姓名：85，年龄：***例外：Prelude.read：无解析。

Answer 1

如果您在请求所有输入后才进行验证，我只会使用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

请注意我们如何在return语句中使用Maybe monad。

如果你想在输入无效值时退出请求输入，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，则不会要求他们提供性别。

MaybeT / Maybe和IO：故障安全阅读信息

1 个答案: