我可以在 Haskell 的守卫中使用守卫吗答案

【问题标题】：Can I use guards within guards in Haskell我可以在 Haskell 的守卫中使用守卫吗
【发布时间】：2023-09-13 12:22:01
【问题描述】：

我在 Haskell 的守卫中尝试使用守卫时遇到了一个问题，我试图找出是否（不可能）这样做，但我在互联网上找不到任何答案，我希望你小伙伴们可能知道答案了。这是我在尝试编译代码时遇到的错误： Assignment2.hs:134:75: parse error on input '|'

我的代码：

    verticals :: Board -> (Row, Row, Row)   
    verticals ((a,b,c),(d,e,f),(g,h,i)) = ((a,d,g),(b,e,h),(c,f,i))

    symbolToPlayer :: Field -> Player   
    symbolToPlayer X = P1  
    symbolToPlayer O = P2

    showTime :: Board -> Maybe Player  
    showTime ((a,b,c),(d,e,f),(g,h,i))
                        | a==b && a==c = Just (symbolToPlayer a)          
                        | d==e && d==f = Just (symbolToPlayer d)  
                        | g==h && g==i = Just (symbolToPlayer i)
                        | otherwise = False

    hasWinner :: Board -> Maybe Player  
    hasWinner b@((a,b,c),(d,e,f),(g,h,i))       
                                   | showTime b        -> spot of error
                                   | showTime (verticals b)
                                   | a==e && a==i = Just (symbolToPlayer a)
                                   | c==e && c==g = Just (symbolToPlayer c)
                                   | otherwise = Nothing

代码是一个井字游戏的代码，这部分代码应该找出哪个玩家赢了；类型Board自定义，输入示例：hasWinner((X,O,X),(X,O,O),(X,X,O))（输出应为：Just P1）。

感谢您的帮助！

【问题讨论】：

您是否检查过所有的空格都是空格而不是制表符？该函数的获胜者也是->而不是=
我想知道您为什么希望它编译，以及它应该如何工作。显然你想以某种方式评估showTime 的Maybe 结果？

标签： haskell guard

【解决方案1】：

您的程序中至少有 2 个严重的语法错误和一个类型错误。

语法错误：

    | showTime b        -> spot of error
    | showTime (verticals b)

这里缺少右手边。一般形式是

    | guard = expression

没有表情就无法逃脱。编译器等待'='，但找到'|'下一个守卫的，所以它知道自己有问题。

类型错误在这里：

    | g==h && g==i = Just (symbolToPlayer i)
    | otherwise = False

您必须决定函数应该返回Bool 还是Maybe Player。

【讨论】：

【解决方案2】：

这个怎么样？

import Control.Monad (msum)
import Control.Applicative ((<*>), pure)

data Player = P1 | P2 | None deriving (Eq, Show)

data Field = X | O | B deriving (Eq, Show)

type Board = ((Field, Field, Field)
             ,(Field, Field, Field)
             ,(Field, Field, Field))

symbolToPlayer :: Field -> Player
symbolToPlayer X = P1
symbolToPlayer O = P2
symbolToPlayer B = None

checkThree :: (Field,Field,Field) -> Maybe Player
checkThree (a,b,c)
    | a == b && a == c = Just $ symbolToPlayer a
    | otherwise        = Nothing

winHorizontal :: Board -> Maybe Player
winHorizontal (r1, r2, r3) = msum $ map checkThree [r1, r2, r3]

winVertical :: Board -> Maybe Player
winVertical ((a,b,c), (d,e,f), (g,h,i)) =
    msum $ map checkThree [(a,d,g), (b,e,h), (c,f,i)]

winDiagonal :: Board -> Maybe Player
winDiagonal ((a,_,c), (_,e,_), (g,_,i)) =
    msum $ map checkThree [(a,e,i), (c,e,g)]

hasWinner :: Board -> Maybe Player
hasWinner b = msum $ [winHorizontal, winVertical, winHorizontal] <*> pure b

Monad 和 Applicative 是你的朋友！

【讨论】：