Haskell 头/尾与模式匹配

Question

这里有两段代码。

工作:

joins :: [String] -> String -> String
joins [] _ = ""
joins [x] _ = x
joins xs d = head xs ++ d ++ (joins (tail xs) d)

不工作:

joins :: [String] -> String -> String
joins [] _ = ""
joins [x] _ = x
joins [x:xs] d = x ++ d ++ (joins xs d)

后者的错误日志是:

test.hs:4:18:
Couldn't match expected type `[Char]' with actual type `Char'
In the first argument of `(++)', namely `x'
In the expression: x ++ d ++ (joins xs d)
In an equation for `joins':
    joins [x : xs] d = x ++ d ++ (joins xs d)

test.hs:4:35:
Couldn't match type `Char' with `[Char]'
Expected type: [String]
  Actual type: [Char]
In the first argument of `joins', namely `xs'
In the second argument of `(++)', namely `(joins xs d)'
In the second argument of `(++)', namely `d ++ (joins xs d)'

我在这里错过了什么？

Answer 1

使用圆括号，而不是方括号:

   -- vvvvvv
joins (x:xs) d = x ++ d ++ (joins xs d)

模式 [x:xs] 仅匹配长度为 1 的列表，其单个元素是非空列表 x:xs。

因为你的是一个字符串列表，[x:xs] 与 ["banana"] 匹配(其中 x='b', xs= "anana")，使用 ["a"] (x='a', xs="") 但不使用 ["banana ", "split"] 也不与 [""] 一起使用。

这显然不是您想要的，因此请使用普通括号。

(顺便说一句，...++ (joins xs d) 中的括号是不需要的:函数应用程序比 Haskell 中的任何二元运算符绑定(bind)更多。)