Question

嘿，我是Haskell的新手并试图找出如何返回长度为n的单词的列表

getWords :: Int -> [Word] -> [Word]
getWords n w = filter (\x -> length x == n) w

我发现我可以在Prelude中使用它过滤器（\ x - ＆gt;长度x == 5）[“你好”，“23”]

它将返回[“Hello”]，但是当我尝试在函数getWords中执行它时它会给我一个错误

* Couldn't match type `t0 a0' with `Word'
      Expected type: [Word]
        Actual type: [t0 a0]
    * In the expression: filter (\ x -> length x == n) w
      In an equation for `getWords':
          getWords n w = filter (\ x -> length x == n) w
    |
163 | getWords n w = filter (\x -> length x == n) w
    |                ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

Date.hs:163:45: error:
    * Couldn't match type `Word' with `t0 a0'
      Expected type: [t0 a0]
        Actual type: [Word]
    * In the second argument of `filter', namely `w'
      In the expression: filter (\ x -> length x == n) w
      In an equation for `getWords':
          getWords n w = filter (\ x -> length x == n) w
    |
163 | getWords n w = filter (\x -> length x == n) w

我在这里做错了什么？

Answer 1

你做错了是假设Word来自Data.Word。这是一个机器字，如无符号整数值。它不是一个人类的单词，就像字符串中的字母一样。正如@chi的评论中所述，您应该使用[String]：

getWords :: Int -> [String] -> [String]
getWords n w = filter (\x -> length x == n) w

返回长度为n的单词列表的函数

1 个答案: