String Haskell：返回在&；之前的字符串；在给定字符串之后

给定类型字符串，则应返回字符串 给定字符串“Second”应该返回-->“First”，“Third” 这就是这个字符串前面和后面的内容。（某种意义上的对称关系）

以下是我到目前为止所做的尝试

between :: String ->  (String, String)
between "Second" = (“First”,"Third")
between "Third" = (“Second”,"Fourth")
between "Fourth" = (“Third”, "Fifth")

我试过的另一个

data Positions =  First | Second | Third | Fourth | Fifth |

between :: Positions ->  (String,String)
between Second = (“First”,"Third")
between Third = (“Second”,"Fourth")
between Fourth = (“Third”, "Fifth")

接收到不同类型的错误，例如未定义的变量“？”、不正确终止的字符串和公式给出不同的算术

---

您将如何修复此问题，使其能够正常运行，而不会出现任何错误？

此代码中有两个语法错误。首先，在

位置

类型的末尾有一个额外的

|

。我喜欢以不同的方式编写我的ADT，以便在我出错时更容易看到：

data Positions
    = First
    | Second
    | Third
    | Fourth
    | Fifth

另一个错误是，您似乎在每个第一个元组元素周围复制/粘贴了一些非ascii引号字符到源代码中，这很容易修复：

between :: Positions -> (String, String)
between Second = ("First", "Third")
between Third = ("Second", "Fourth")
between Fourth = ("Third", "Fifth")

但是，我还要指出，在您的类型上使用

派生

子句可以更轻松地完成这项工作：

data Positions
    = First
    | Second
    | Third
    | Fourth
    | Fifth
    deriving (Eq, Enum, Bounded)

between :: Positions -> Maybe (Positions, Positions)
between p =
    if p == minBound || p == maxBound
        then Nothing
        else Just (pred p, succ p)

---

这将使用

Maybe

使函数更安全，如果在第一次调用之间调用

，它不会使程序崩溃。如果您真的想要字符串，您可以
派生（Eq，Enum，Bounded，Show）
并使用
仅仅（Show$pred p，Show$such p）
关于您在问题开始时提出的签名（String->（String，String）），它可以是这样的：

between :: String -> (String, String)
between x = case x of
  "Second" -> ("First","Third")
  "Third"  -> ("Second","Fourth")
  "Fourth" -> ("Third","Fifth")

特殊引号（
“
和
”
）是复制和粘贴问题，还是您在源文件中实际使用了这些引号？