函数中的Haskell getLine，否则

Question

我对Haskell非常陌生，我正在尝试学习一些简单的函数。

我通过这样的函数做出了一些选择：

displayOptions :: Int -> String
displayOptions option
    | option == 0 = "1 - Shop\n2 - Fight Monsters\n3 - Inn\n4 - Monk"
    | otherwise = "invalid"

然后使用getLine获取用户输入。

choice <- getLine

然后显示第二个选项框，例如，

playerChoice :: String -> String
playerChoice option
    | option == "1" = "Sword - 50gp\nShield - 100gp"
    | option == "2" = "You go fight some monsters outside town."
    | option == "3" = "You go to the town Inn."
    | option == "4" = "You go see the holy monk."
    | otherwise = "You entered invalid information...\n" ++ displayOptions(0)

我感到困惑的是如何在函数中再次获得用户的选择。我希望我的otherwise =说出无效的信息，显示选项，再次获得输入，然后显示他们所做的选择。

所以我的主程序应该是这样的：

main = do
    putStrLn "The king has sent you on the journey to become a master."
    putStrLn $ "What would you like to do?"
    putStrLn $ displayOptions(0)
    choice <- getLine
    putStrLn $ playerChoice(choice)

Answer 1

您必须将返回类型更改为IO字符串，而不是字符串。

但是，您可能希望返回任何一个字符串，以便该函数指示它返回了游戏进程文本Right "You do something"或一个失败，并解释了失败Left "Not an option"。

然后在调用者中循环，直到得到一个Right值，每次得到一个Left值，就会打印文本并再次询问。

我确信有一个稍微好一点的方法，但下面是一些快速修复代码：

module Main where

playerChoice :: String -> Either String String
playerChoice option
    | option == "1" = Right "Sword - 50gp\nShield - 100gp"
    | option == "2" = Right "You go fight some monsters outside town."
    | option == "3" = Right "You go to the town Inn."
    | option == "4" = Right "You go see the holy monk."
    | otherwise = Left "You entered invalid information..."

displayOptions :: Int -> String
displayOptions option
    | option == 0 = "1 - Shop\n2 - Fight Monsters\n3 - Inn\n4 - Monk\n"
    | otherwise = "invalid"

main = do
    let progress whathappens = do
        putStrLn whathappens
        let tryAsk prompt = do
            putStrLn prompt
            choice <- getLine
            either tryAsk progress $ playerChoice(choice)
        tryAsk $ displayOptions(0) ++ "What would you like to do?"
    progress "The king has sent you on the journey to become a master."

如果您是import Data.Function，那么您也可以像下面这样编写它--在本例中可能不是更好，但这是进入haskell迷人部分的一个不错的浅表步骤：

fix (\moreProgress whathappens -> do
    putStrLn whathappens
    fix (\askAgain prompt -> do
        putStrLn prompt
        choice <- getLine
        either askAgain moreProgress $ playerChoice(choice))
        $ displayOptions(0) ++ "What would you like to do?")
    $ "The king has sent you on the journey to become a master."

Answer 2

1. 您应该将返回类型更改为Either String String，以便提供错误消息。有关更多细节，请查看Either类型的文档。
2. 在Haskell中，我们没有像for或while这样的传统循环结构，而是使用递归调用。有关示例，请参见While loop in Haskell with a condition。