Question

我真的是Haskell的新人。对于主编码器来说，这也许是一个简单的问题。我想删除字符串外的引号。例如“ A1”到A1。

我尽力解决了这个问题。但这不起作用。我已经使用过read :: Read a => String-> a，id函数和正则表达式。

initialGuess :: ([Pitch],GameState)
initialGuess = (startGuess, GameState (chords))
      where startGuess = ["A1", "B1", "C1"] 
      standard = [[n, o] | n <- ['A'..'G'], o <- ['1'..'3']]  <br/>
      chords = [[a,b,c] | a <- standard, b <- standard, c <-  
               standard, a /= b, b /= c, a /= c]

initialGuess的目的是不接受任何输入参数，并返回一对初始猜测值和一个游戏状态。运行此代码，我可以得到

[“ A1”，“ B1”，“ C1”]，GameState [[“ A1”，“ A2”，“ A3”]，[“ A1”，“ A2”，“ B1”]，[“ A1 “，” A2“，” B2“]，[” A1“，” A2“，” B3“]，[” A1“，” A2“，” C1“] ............ .... [“ A1”，“ A2”，“ C2”]，[“ A1”，“ A2”，“ C3”]，[“ A1”，“ A2”，“ D1”]，[“ A1” ，“ A2”，“ D2”]，[“ A1”，“ A2”，“ D3”]]

但是，我想删除这些引号，例如

[A1，B1，C1]，GameState [[A1，A2，A3]，[A1，A2，B1]，[A1，A2，B2]，[A1，A2，B3]，[A1，A2， C1] ............. [A1，A2，C2]，[A1，A2，C3]，[A1，A2，D1]，[A1，A2，D2]， [A1，A2，D3]]

Answer 1

如果您只是想让状态更漂亮，则可以提供自己的显示功能：

-- |Display a list of strings without quotation marks
showLStr :: [String] -> String
showLStr p  = "[" ++ intercalate "," p ++ "]"

intercalate函数将","放在字符串列表的每个元素之间。

showLStr ["A1", "A2", "A3"]

应显示为

[A1,A2,A3]

编辑：现在您可以使用showLStr来显示游戏状态：

showGameState :: GameState -> String
showGameState (GameState chords) =
  "GameState [" ++ (intercalate "," $ map showLStr chords) ++ "]"

-- |Make GameState "Show"-able
instance Show GameState where
  show = showGameState

showGuess :: ([Pitch],GameState) -> String
showGuess (pitch, gameState) = showLStr pitch ++ ", " ++ show gameState

如何删除字符串中的引号

1 个答案: