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Haskell-在用其中一个验证后继续执行下一个操作

either scope haskell validation

Jamin · 技术社区 · 6 年前

我是Haskell的新手,使用do符号来验证用户的选择。

userChoice :: String -> Either String String
userChoice option
    | option == "1" = Right "You proceed with option 1."
    | option == "2" = Right "You proceed with option 2"
    | option == "3" = Right "You proceed with option 3"
    | option == "4" = Right "You proceed with option 4"
    | otherwise = Left $ "\n\nInvalid Option...\nWhat would you like to do?\n" ++ displayOptions(0)

displayOptions :: Int -> String
displayOptions option
    | option == 0 = "1 - Option 1\n2 - Option 2\n3 - Option 3\n4 - Option 4"
    | otherwise = "invalid"

main = do
    putStrLn "Welcome."
    let start startMessage = do
        putStrLn startMessage
        let ask message = do
            putStrLn message
            choice <- getLine
            either ask ask $ userChoice(choice)
        ask $ "What would you like to do?\n" ++ displayOptions(0)
    start "Choose a number between 1-4."

这很好,但是在用户选择了正确的选项之后,我想继续程序的下一部分。我可以使用return,但是我失去了用户选择的选项。

main = do
    putStrLn "Welcome."
    let start startMessage = do
        putStrLn startMessage
        let ask message = do
            putStrLn message
            choice <- getLine
            either ask return $ userChoice(choice)
        ask $ "What would you like to do?\n" ++ displayOptions(0)
    start "Choose a number between 1-4."

    -- putStrLn userChoice2(choice)

    let continue continueMessage = do
        putStrLn continueMessage
        let ask message = do
            putStrLn message
            choice <- getLine
            either continue continue $ userChoice(choice)
        ask $ "What would you like to do?"
    continue "We are now here..."

如果我尝试将continue作为正确的选项,那么它会抛出一个范围错误。类似地,如果我使用return并尝试为userChoice创建一个原始的克隆函数,我将无法再访问choice所在的范围。

userChoice2 :: String -> String
userChoice2 option
    | option == "1" = "You proceed with option 1."
    | option == "2" = "You proceed with option 2"
    | option == "3" = "You proceed with option 3"
    | option == "4" = "You proceed with option 4"
    | otherwise = "\n\nInvalid Option...\nWhat would you like to do?\n" ++ displayOptions(0)

1 回复 | 直到 6 年前

jberryman 6 年前

听起来您只需要绑定monad中返回的值 start ,如下所示:

main = do
    putStrLn "Welcome."
    let start startMessage = do
        putStrLn startMessage
        let ask message = do
            putStrLn message
            choice <- getLine
            either ask return $ userChoice(choice)
        ask $ "What would you like to do?\n" ++ displayOptions(0)

    -- THE NEW BIT I ADDED:
    opt <- start "Choose a number between 1-4."
    putStrLn opt   -- THIS PRINTS "You proceed with ..."

    -- SNIP

你可以读到关于“脱糖”的书 do 符号”来学习如何将其转换为lambdas和 Monad 班级( >>= 和 return ).

您可以使用(而且应该更喜欢)模式匹配,而不是保护和 == Eq 类,而不是所有类型都实现)。例如。

userChoice2 option = case option of
    "1" -> "You proceed with option 1."
    "2" -> "You proceed with option 2"

记住在haskell中调用 foo 在 a 和 b foo a b 不 foo(a,b)

打开警告是非常有帮助的(无论是在学习时还是在生产代码基础上),例如,在这里,我在打开警告后在ghci中加载您的文件 -Wall

Prelude> :set -Wall
Prelude> :l k.hs
[1 of 1] Compiling Main             ( k.hs, interpreted )

k.hs:15:1: warning: [-Wmissing-signatures]
    Top-level binding with no type signature: main :: IO b
   |
15 | main = do
   | ^^^^

k.hs:24:5: warning: [-Wunused-do-bind]
    A do-notation statement discarded a result of type âStringâ
    Suppress this warning by saying
      â_ <- start "Choose a number between 1-4."â
   |
24 |     start "Choose a number between 1-4."
   |     ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Ok, one module loaded.

注意它警告你 开始 IO String String 值,这确实是一个错误。如果你真的想放弃你可以使用的值 void