此代码导致堆栈溢出错误 - 你们中的任何人都可以看到我错过的可能导致它的任何内容吗？我已经完成了所有的功能，并将它们设置为只返回任意值，但堆栈溢出错误仍然显示出来..
module Reversi where
import Data.List
-- Position type and utility functions
type Position = (Int, Int)
-- Given a Position value, determine whether or not it is a legal position on the board
isValidPos :: Position -> Bool
isValidPos (a,b)
| a > 8 || b > 8 = False
| a < 1 || b < 1 = False
| otherwise = True
-- Player type and utility functions
data Player = PlayerWhite | PlayerBlack deriving (Eq)
instance Show Player where
show PlayerWhite = "white"
show PlayerBlack = "black"
-- Given a Player value, return the opponent player
otherPlayer :: Player -> Player
otherPlayer a
| a == PlayerWhite = PlayerBlack
| otherwise = PlayerWhite
-- Piece type and utility functions
data Piece = Piece Position Player deriving (Eq)
instance Show Piece where
show (Piece _ PlayerWhite) = " W"
show (Piece _ PlayerBlack) = " B"
-- Given a Player value and a Piece value, does this piece belong to the player?
isPlayer :: Player -> Piece -> Bool
isPlayer a (Piece (x,y) z)
| a == z = True
| otherwise = False
-- Given a Piece value, determine who the piece belongs to
playerOf :: Piece -> Player
playerOf a
| show a == " W" = PlayerWhite
| otherwise = PlayerBlack
-- Flip a piece over
flipPiece :: Piece -> Piece
flipPiece (Piece (x,y) z)
| z == PlayerWhite = (Piece (x,y) PlayerBlack)
| otherwise = (Piece (x,y) PlayerWhite)
-- Board type and utility functions
type Board = [Piece]
-- The initial configuration of the game board
initialBoard :: Board
initialBoard =
[
Piece (3,4) PlayerWhite, Piece (4,4) PlayerBlack,
Piece (3,3) PlayerBlack, Piece (4,3) PlayerWhite
]
-- Given a Position value, is there a piece at that position?
isOccupied :: Position -> Board -> Bool
isOccupied (x,y) b
| any (\(Piece b c) -> b == (x,y)) b = True
| otherwise = False
-- Which piece is at a given position?
-- Return Nothing in the case that there is no piece at the position
-- Otherwise return Just the_piece
pieceAt :: Position -> Board -> Maybe Piece
pieceAt (x,y) b
| isOccupied (x,y) b = (find (\(Piece (a,b) player) -> (a,b) == (x,y)) b)
| otherwise = Nothing
-- ***
-- Determine if a particular piece can be placed on a board.
-- There are two conditions:
-- (1) no two pieces can occupy the same space, and
-- (2) at least one of the other player's pieces must be flipped by the placement of the new piece.
validMove :: Piece -> Board -> Bool
validMove (Piece (x,y) p) b
| isOccupied (x,y) b && toFlip (Piece (x,y) p) b /= [] = True
| otherwise = False
-- ***
-- Determine which pieces would be flipped by the placement of a new piece
toFlip :: Piece -> Board -> [Piece]
toFlip (Piece (x,y) player) b
| validMove (Piece (x,y) player) b = (getLineDir (-1,-1) (Piece (x,y) player) b) ++
(getLineDir (-1,0) (Piece (x,y) player) b) ++
(getLineDir (-1,1) (Piece (x,y) player) b) ++
(getLineDir (0,-1) (Piece (x,y) player) b) ++
(getLineDir (0,0) (Piece (x,y) player) b) ++
(getLineDir (0,1) (Piece (x,y) player) b) ++
(getLineDir (1,-1) (Piece (x,y) player) b) ++
(getLineDir (1,0) (Piece (x,y) player) b) ++
(getLineDir (1,1) (Piece (x,y) player) b)
| otherwise = []
-- ***
-- Auxillary function for toFlip.
-- You don't have to use this function if you prefer to define toFlip some other way.
-- Determine which pieces might get flipped along a particular line
-- when a new piece is placed on the board.
-- The first argument is a vector (pair of integers) that describes
-- the direction of the line to check.
-- The second argument is the hypothetical new piece.
-- The return value is either the empty list,
-- a list where all pieces belong to the same player,
-- or a list where the last piece belongs to the player of the hypothetical piece.
-- Only in the last case can any of the pieces be flipped.
getLineDir :: (Int, Int) -> Piece -> Board -> [Piece]
getLineDir (x1,y1) (Piece (x,y) player) b
| isOccupied (x*x1, y*y1) b && (pieceAt (x, y) b) == (pieceAt (x*x1, y*y1) b) = (concat . concat) (map (\(Piece (x,y) player) -> drop 1 [filter (\(Piece (x,y) player) -> (x,y) == (x*x1, y*y1)) b]++[(getLineDir (x*x1, y*y1) (Piece (x,y) player) b)]) b)
| otherwise = []
--getLineDir :: (Int, Int) -> Piece -> Board -> [Piece]
--getLineDir (x1,y1) (Piece (x,y) player) b
-- | isOccupied (x*x1, y*y1) b && (pieceAt (x, y) b) == (pieceAt (x*x1, y*y1) b) = (Piece (x, y) player):(getLineDir (x*x1, y*y1) (Piece (x,y) player) b)
-- | otherwise = []
-- ***
-- Auxillary function for toFlip.
-- You don't have to use this function if you prefer to define toFlip some other way.
-- Given the output from getLineDir, determine which, if any, of the pieces would be flipped.
--flippable :: [Piece] -> [Piece]
-- ***
-- Place a new piece on the board. Assumes that it constitutes a validMove
makeMove :: Piece -> Board -> Board
makeMove p b = [p]++b
-- ***
-- Find all valid moves for a particular player
allMoves :: Player -> Board -> [Piece]
allMoves p ((Piece (x,y) plr):bs) =
if p == plr then ( toFlip (Piece (x,y) p) [Piece (x,y) p]++bs )++(allMoves p bs) else []
--allMoves :: Player -> Board -> [Piece]
--allMoves p b
-- | filter (\(Piece (x,y) player) -> player == p) b /= [] = (concat . concat) (map (\x -> drop 1 [filter (\(Piece (x,y) player) -> player == p) b]++[(toFlip x b)] ) (filter (\(Piece (x,y) player) -> player == p) b))
-- | otherwise = []
-- ***
-- Count the number of pieces belonging to a player
score :: Player -> Board -> Int
score player [] = 0 score player ((Piece pos plr):bs) = if player == plr then 1 + score player bs else score player bs
-- Decide whether or not the game is over. The game is over when neither player can make a validMove
isGameOver :: Board -> Bool
isGameOver b
| allMoves PlayerBlack b == [] && allMoves PlayerWhite b == [] = True
| otherwise = False
-- Find out who wins the game.
-- Return Nothing in the case of a draw.
-- Otherwise return Just the_Player
winner :: Board -> Maybe Player
winner b
| score PlayerWhite b > score PlayerBlack b = Just PlayerWhite
| score PlayerWhite b < score PlayerBlack b = Just PlayerBlack
| otherwise = Nothing
答案 0 :(得分：3)
这可能是由一个永远不会结束的递归引起的。您可能想要进一步深入getLineDir进行递归，并且不清楚它是否正在完成。
您可能还想使用调试器来查看代码的执行情况（请参阅http://www.haskell.org/ghc/docs/6.10.4/html/users_guide/ghci-debugger.html）。
此外，您的代码可以简化。每次你有类似的东西：
f :: a -> Bool
f a | someGuard = True
| otherwise = False
你可以用
替换它f :: a -> Bool
f a = someGuard
另一件事，[x]++xs最好这样做：x : xs。 （见makeMove）
最后的改变：
flipPiece :: Piece -> Piece
flipPiece (Piece (x,y) z)
| z == PlayerWhite = (Piece (x,y) PlayerBlack)
| otherwise = (Piece (x,y) PlayerWhite)
可以像这样简化，例如，
flipPiece :: Piece -> Piece
flipPiece (Piece xy PlayerWhite) = Piece xy PieceBlack
flipPiece (Piece xy PlayerBlack) = Piece xy PieceWhite
答案 1 :(得分：0)
为什么要增加方向和位置？我想你需要添加它们！
如果您想检查(5,4)上的展示位置是否对Black有效，您的代码将尝试计算getLineDir (-1,-1) (Piece (5,4) PieceBlack) board。将要计算的第一件事是isOccupied (-5,-4) board，这对我来说似乎不对 - 但如果你用加法替换乘法，那么isOccupied (4,3) board将被调用。
正如现在的代码所示，getLineDir中有很多地方通过方向抵消位置：
getLineDir (x1,y1) (Piece (x,y) player) b
| isOccupied (x*x1, y*y1) b && (pieceAt (x, y) b) == (pieceAt (x*x1, y*y1) b) = (concat . concat) (map (\(Piece (x,y) player) -> drop 1 [filter (\(Piece (x,y) player) -> (x,y) == (x*x1, y*y1)) b]++[(getLineDir (x*x1, y*y1) (Piece (x,y) player) b)]) b)
| otherwise = []
我建议将这种逻辑分解为类似
的东西type Direction = (Int,Int)
offset :: Postion -> Direction -> Position
offset (x,y) (deltax,deltay) = (x+deltax, y+deltay)
然后getLineDir变为（未经测试！）：
getLineDir dir (Piece pos player) b
| isOccupied (offset pos dir) b && pieceAt pos b == pieceAt (offset pos dir) b = (concat . concat) (map (\(Piece (x,y) player) -> drop 1 [filter (\(Piece (x,y) player) -> (x,y) == offset (x,y) dir) b]++[(getLineDir (offset (x,y) dir) (Piece (x,y) player) b)]) b)
| otherwise = []
where newPos = offset pos dir
我还建议重命名你在内部lambda中绑定的变量，很难看出哪个x，y和player是哪个！但这可能更适合Codereview Stackexchange！