Haskell中的一个简单的绞刑游戏

Question

我刚刚完成了大学的函数式编程课程，并继续学习Haskell，因为我发现它非常有趣！我做了一个简单的绞刑架游戏，并希望听到您的任何想法和想法，可能会出现在代码！例如，在函数式编程的意义上，怎样才能使它更加教条主义呢？

import Control.Monad
import Data.List (elemIndices, sort)

pictures = 9

main :: IO ()
main = do
  word <- getWord ""
  clearScreen
  hang word

hang :: String -> IO ()
hang word = hang' word [] [] pictures

hang' :: String -> [Char] -> [Char] -> Int -> IO ()
hang' word _ _ lives | lives == 0 = clearScreen >> putStrLn (renderHangman 0) >> putStrLn "You lost!" >> putStrLn ("The correct word was " ++ word ++ "\n")
hang' word rights _ lives | win rights word = clearScreen >> putStrLn (renderHangman lives) >> putStrLn "You won!" >> putStrLn ("The correct word was " ++ word ++ "\n")
hang' word rights wrongs lives = do
  clearScreen
  putStrLn $ renderHangman lives
  putStrLn $ renderWord rights word
  putStrLn $ renderLives lives
  putStrLn $ renderWrongs wrongs
  guess <- getGuess
  if guess `elem` (rights ++ wrongs)
    then hang' word rights wrongs lives
    else
      if correctGuess guess word
        then hang' word (guess : rights) wrongs lives
        else hang' word rights (guess : wrongs) (lives - 1)

win :: [Char] -> String -> Bool
win guesses = all (`elem` guesses)

clearScreen :: IO ()
clearScreen = replicateM_ 100 (putStrLn "")

correctGuess :: Char -> String -> Bool
correctGuess guess word = guess `elem` word

getGuess :: IO Char
getGuess = do
  putStrLn "Guess a letter!"
  getChar

getWord s = do
  clearScreen
  putStrLn "Give a secret word!"
  putStr ['*' | _ <- s]
  c <- getChar
  case c of
    '\n' -> return s
    char -> getWord (s ++ [char])

renderWord :: [Char] -> String -> String
renderWord guesses = foldr hide ""
  where
    hide = \x xs -> if x `elem` guesses then x : xs else '_' : xs

renderWrongs :: [Char] -> String
renderWrongs [] = ""
renderWrongs wrongs = "Wrong guesses: " ++ sort wrongs

renderHangman :: Int -> String
renderHangman = unlines . hangmanpics

renderLives :: Int -> String
renderLives lives = show lives ++ " guesses left!"

hangmanpics :: Int -> [String]
hangmanpics 9 = ["       ", "       ", "       ", "       ", "       ", "       ", "========="]
hangmanpics 8 = ["       ", "      |", "      |", "      |", "      |", "      |", "========="]
hangmanpics 7 = ["  +---+", "      |", "      |", "      |", "      |", "      |", "========="]
hangmanpics 6 = ["  +---+", "  |   |", "      |", "      |", "      |", "      |", "========="]
hangmanpics 5 = ["   +---+", "   |   |", "   O   |", "       |", "       |", "       |", " ========="]
hangmanpics 4 = ["   +---+", "   |   |", "   O   |", "   |   |", "       |", "       |", " ========="]
hangmanpics 3 = ["   +---+", "   |   |", "   O   |", "  /|   |", "       |", "       |", " ========="]
hangmanpics 2 = ["   +---+", "   |   |", "   O   |", "  /|\\  |", "       |", "       |", " ========="]
hangmanpics 1 = ["   +---+", "   |   |", "   O   |", "  /|\\  |", "  /    |", "       |", " ========="]
hangmanpics 0 = ["   +---+", "   |   |", "   O   |", "  /|\\  |", "  / \\  |", "       |", " ========="]
hangmanpics _ = ["   +---+", "   |   |", "   O   |", "  /|\\  |", "  / \\  |", "       |", " ========="]
```#qcStackCode#

Answer 1

看上去真不错！

以下是一些建议，按意义的大致顺序排列

1. 游戏本质上可以用三个值( word、rights、wrongs和lives )来描述，这三个值在函数之间传递。为了可读性，我建议将它们包装成一个数据类型: data GameState = GameState { word :：String，right：，lives :：Int }
2. hang'的结束让人感到困惑。我所说的“结束”是指getGuess之后的代码。语义结构是：“更新”“游戏状态”，然后重新启动hang'。但是，用当前的代码结构很难看出这一点，因为每个分支都是对hang'的“独立”调用。将状态存储在自己的GameState类型中，我们可以将代码清晰地拆分为以下部分：(1)更新状态；(2)调用hang'。接下来，它看起来是这样的:猜测<- getGuess let state‘= if猜测elem (权利状态++错误状态)，然后声明correctGuess猜测状态，如果correctGuess猜测状态，则状态{ rights =猜测: rights } lives { if =猜测:错误状态，生命=生命状态-1}if’state‘state。
3. [Char]不是rights和wrongs的适当数据类型。在这里使用[Char]向我表明，您关心Chars在列表中的顺序，以及他们出现了多少次，但实际上您不关心。相反，我推荐Data.Set。(这也会给你更好的表现！)
4. getWord可以更简单地实现为clearScreen >> putStrLn "Give a secret word!" >> getLine
5. win --这个函数的圆滑实现！不相关，我会重命名它。win听起来像一个动作，但函数实际上是一个谓词。也许是isWin？
6. 作为一般的经验法则，我会尽量避免像import Control.Monad这样的不合格的进口产品。如果您有多个不合格的导入，则很难知道从哪里导入函数或操作符。
7. getWord缺少它的类型签名，pictures也是如此。这并不是什么大问题，但通常建议在(至少)所有顶级值上包含类型签名。
8. 在某些地方，您使用>>链接D42操作，而在其他地方，则使用do符号。这实际上是一回事；thingOne >> thingTwo和do { thingOne; thingTwo; }是相同的代码。(我不清楚你是否已经知道了)无论如何，为了可读性，我建议使用do来表示更长的链，而使用>>或>>=来表示更短的表达式。具体而言，使用clearScreen >> putStrLn ...重写hang'的do部分，然后将getGuess重写为putStrLn "Guess a letter!" >> getChar。

希望这会有所帮助:-)查看所有更改的代码(Set除外) 这里