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

使用括号，而不是方括号：

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

模式[x:xs]只匹配长度为1的列表，其单个元素是一个非空列表x:xs。

由于您的是字符串列表，[x:xs] 与 ["banana"]（其中 x='b', xs="anana"）匹配，与 ["a"]（x='a', xs=""）匹配，但与 ["banana", "split"] 不匹配，也不与 [""] 匹配。

这显然不是你想要的，所以使用简单的括号。

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