haskell - 如何通过检查是否满足获胜条件来退出 Haskell 中的 IO？

Question

我正在尝试在 Haskell 中实现连接四的游戏。此时它基本上可以正常工作，但我很难将游戏已经结束的事实传递给 IO 递归函数。目前，代码将声明我有“函数游戏循环中的非详尽模式”，并且我不知道在实现胜利状态时如何正确退出。

这里是一些代码片段:

data Action = Move Piece State Col   -- do Piece in State
            | Start              -- returns starting state
data Result = EndOfGame Int         -- end of game
            | ContinueGame State
         deriving (Eq, Show)

type Game = Action -> Result


connectfour :: Game
connectfour Start = ContinueGame ([],[],[],[],[],[],[])
connectfour (Move move state col)
    | win move (addPiece move state col) col = EndOfGame move -- player of Piece value move wins
    | isTie (addPiece move state col) = EndOfGame 0 -- no moves, tie
    | otherwise = ContinueGame (addPiece move state col)


gameLoop (ContinueGame state) player = do
  let turn = if player == 1 then 2 else 1
  -- Removed code for forum, just prepping values to feed into printboard and printing dialog
  putStrLn "Please select a column to enter your piece"
  input <- getLine
  gameLoop (connectfour (Move turn state (read input :: Int))) turn

TDLR； 我试图返回一个字符串，说明哪个玩家在退出时获胜，但我不知道如何编写一个在游戏循环中检查此情况的条件。

Answer 1

您只需在 gameLoop 中添加 EndOfGame 的情况即可:

gameLoop :: Result -> Int -> IO ()
gameLoop (EndOfGame move) _ = putStrLn ("Game over! Player " ++ show move ++ " won")
gameLoop (ContinueGame state) player = do
  {- omitted -}
  gameLoop (connectfour (Move turn state (read input :: Int))) turn

“函数 gameloop 中的非详尽模式”警告告诉您 gameLoop 缺少一个案例 - EndOfGame 案例。

---

另一方面，您可能需要考虑制作一个播放器 ADT:

data Player = One | Two deriving (Show)

otherPlayer :: Player -> Player
otherPlayer One = Two
otherPlayer Two = One