Question

给定一个String类型，应返回Strings 例如。给定字符串“Second”应返回 - ＆gt; “第一”，“第三” 这就是这个字符串之前的内容以及之后的内容。（在某种意义上的对称关系）

这是我到目前为止所尝试的内容。

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")

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方程式给出了不同的元素。

如何解决这个问题以使其正常运行，没有任何错误？

Answer 1

此代码中有两个语法错误。首先，您在|类型的末尾有一个额外的Positions。我喜欢以不同的方式编写ADT，以便在我搞砸之时更容易看到：

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

另一个错误是你似乎已经将一些非ascii引号字符复制/粘贴到每个第一个元组元素的源代码中，并且很容易修复：

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deriving子句可以更轻松地完成此操作：

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)

这使用Maybe来使函数更安全，如果您调用between First，它不会导致程序崩溃。如果您真的想要字符串，则可以derive (Eq, Enum, Bounded, Show)使用Just (show $ pred p, show $ succ p)

Answer 2

尊重您在问题开头提出的签名（String - ＆gt;（String，String）），它看起来像这样：

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

Haskell：返回之前和之前的字符串在给定的字符串之后

2 个答案: