如何在Haskell中从构造方法转换为字符串？

Question

我正在尝试以字符串格式输出一手牌，但是我不能从牌的等级中删除单词“ Numeric”，我宁愿不使用辅助函数来执行此操作，因为它看起来确实很乱，任何建议在没有帮助功能的情况下执行此操作。另外，当我使用display功能时，我在前奏中使用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

我宁愿不使用辅助功能也要这样做，因为它看起来确实很凌乱

假设您具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)

然后编写辅助函数会很有意义

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

displaySuit :: Suit -> String
displaySuit = show

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

但是您也可以将它们组合：

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

具有正确设计的辅助函数并不麻烦。相反。

Answer 2

您用show (rank c)完成的工作，

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

您可以用displayRank (rank c)完成

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

（谈到类型）。