Haskell数据域中的多态性_Haskell_Polymorphism_Field_Chess_Maybe

Haskell数据域中的多态性

haskell

Haskell数据域中的多态性,haskell,polymorphism,field,chess,maybe,Haskell,Polymorphism,Field,Chess,Maybe,我试图通过创建一个简单的国际象棋游戏来学习Haskell。但是，我在定义表示电路板正方形的函数时遇到了问题 import Data.Char type Board = [[Square]] type Square = Maybe Piece data Piece = Piece PieceColor PieceType deriving (Show, Eq) data PieceColor = White | Black deriving (Show, Eq) data PieceType =

我试图通过创建一个简单的国际象棋游戏来学习Haskell。但是，我在定义表示电路板正方形的函数时遇到了问题

import Data.Char
type Board = [[Square]]
type Square = Maybe Piece
data Piece = Piece PieceColor PieceType deriving (Show, Eq)
data PieceColor = White | Black deriving (Show, Eq)
data PieceType = King | Queen | Rook | Bishop | Knight | Pawn deriving (Show, Eq)

...    

displaySquare :: Square -> Char
    displaySquare n
        | n == Nothing = ' '
        | n == Just (Piece White _) = displaySquare' n
        | otherwise = toLower (displaySquare' n)
            where
                displaySquare'   (Just (Piece _ King))   = 'K'
                displaySquare'   (Just (Piece _ Queen))  = 'Q'
                displaySquare'   (Just (Piece _ Rook))   = 'R'
                displaySquare'   (Just (Piece _ Bishop)) = 'B'
                displaySquare'   (Just (Piece _ Knight)) = 'N'
                displaySquare'   (Just (Piece _ Pawn))   = 'P'

尝试运行GHCI将返回以下错误：

Chess.hs:21:30:
    Found hole ‘_’ with type: PieceType
    Relevant bindings include
      displaySquare' :: Maybe Piece -> Char (bound at Chess.hs:24:13)
      n :: Square (bound at Chess.hs:19:15)
      displaySquare :: Square -> Char (bound at Chess.hs:19:1)
    In the second argument of ‘Piece’, namely ‘_’
    In the first argument of ‘Just’, namely ‘(Piece White _)’
    In the second argument of ‘(==)’, namely ‘Just (Piece White _)’
Failed, modules loaded: none.

我不太清楚错误是想告诉我什么。如果我不得不猜测的话，我会说问题是将数据字段定义为一个洞“z”，因为这意味着函数可以将任何类型作为数据字段，这显然与给定的类型签名冲突。我的假设正确吗？我应该如何解决它？

这是使用模式匹配编写

displaySquare

的方法：

displaySquare Nothing = ' '
displaySquare n@(Just (Piece White _)) = displaySquare' n
displaySquare n                        = toLower (displaySquare' n)
  where
    ...your definition of displaySquare'...

您在定义

displaySquare'

时正确使用了模式匹配

注意语法

n@（Just（Piece White））

是如何工作的——它提供了一个模式来匹配：

Just White

，并且它还将变量

设置为传递给函数的参数

更新

@behzad.nouri在评论中提到，您需要将helper函数置于顶层

另一种使助手函数保持本地的写入方法：

displaySquare n =
  case n of
    Nothing              -> ' '
    Just (Piece White _) -> displaySquare' n
    _                    -> toLower (displaySquare' n)
  where
    displaySquare' x = ...

在这种情况下，

绑定到

displaySquare

的第一个参数，因此不需要使用

n@...

语法。