Haskell 中如何将构造函数转换为字符串？

Question

我正在尝试以字符串格式输出一手牌，但我不能从牌的等级中删除单词 "Numeric"，我更愿意在没有辅助函数的情况下这样做，因为它看起来真的很乱，任何建议都可以在没有辅助功能的情况下执行此操作。此外，当我使用 display 函数时，我在 Prelude 中使用了 putStr (display x)。这是到目前为止的代码：

displayCard :: Card -> String
displayCard c = show (rank c) ++ " of " ++ show (suit c) ++ "\n"

display :: Hand -> String
display Empty = "\n"
display (Add c h) = displayCard c
                    ++ display h

displayRank :: Rank -> String
displayRank (Numeric 2) = "2"
displayRank (Numeric 3) = "3"
displayRank (Numeric 4) = "4"
displayRank (Numeric 5) = "5"
displayRank (Numeric 6) = "6"
displayRank (Numeric 7) = "7"
displayRank (Numeric 8) = "8"
displayRank (Numeric 9) = "9"
displayRank (Numeric 10) = "10"
displayRank Jack = "Jack"
displayRank Queen = "Queen"
displayRank King = "King"
displayRank Ace = "Ace"

Answer 1

你用 show (rank c)、

完成的事情

rank ::        Card -> Rank
show ::                Rank -> String

你可以用 displayRank (rank c)

完成

rank ::        Card -> Rank
displayRank ::         Rank -> String

（说到类型）。

Answer 2

I would prefer to do this without a helper function because it seems really messy

假设您有类似以下类型的内容，

data Card = Card { cardRank :: Rank, cardSuit :: Suit }
  deriving (Eq, Show)

data Rank = Numeric Int | Jack | Queen | King | Ace
  deriving (Eq, Show)

data Suit = Diamonds | Clubs | Hearts | Spades
  deriving (Eq, Show)

那么编写辅助函数就有意义了

displayRank :: Rank -> String
displayRank (Numeric n) = show n
displayRank otherRank = show otherRank

displaySuit :: Suit -> String
displaySuit = show

displayCard :: Card -> String
displayCard (Card rank suit) = ...

但您也可以将它们组合起来：

displayCard :: Card -> String
displayCard (Card (Numeric n) suit) = ...
displayCard (Card otherRank suit) = ...

正确设计辅助函数并不麻烦。相反。

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