Haskell头部/尾部与花纹匹配

Question

以下是两段代码。

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) 

的错误日志，后者则是：

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)' 

缺少什么我在这里？

Answer 1

使用括号，括号不：

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

与长度为1的列表中，其单个元件是一个非空列表x:xs图案仅[x:xs]匹配。

因为你是一个字符串列表，[x:xs]与["banana"]场比赛（其中x='b', xs="anana"），用["a"]（x='a', xs=""），但与["banana", "split"]也不符合[""]。

这显然不是你想要的，所以使用简单的圆括号。

（顺便说一下，不需要在... ++ (joins xs d)括号：功能应用的优先级比在Haskell任何二进制运算符的更多。）