是否有任何haskell函数将列表与分隔符连接起来？

有没有用分隔符连接列表元素的功能？例如：

> foobar " " ["is","there","such","a","function","?"]
["is there such a function ?"]

感谢您的回复！

是的，有：

Prelude> import Data.List
Prelude Data.List> intercalate " " ["is","there","such","a","function","?"]
"is there such a function ?"

intersperse 比较笼统：

Prelude> import Data.List
Prelude Data.List> concat (intersperse " " ["is","there","such","a","function","?"])
"is there such a function ?"

此外，对于要与空格字符连接的特定情况，还有unwords：

Prelude> unwords ["is","there","such","a","function","?"]
"is there such a function ?"

unlines其工作原理类似，只是使用换行符将字符串内插，并且将换行符也添加到末尾。（这对于序列化文本文件很有用，根据POSIX标准，文本文件必须以尾随换行符结尾）

使用foldr编写单行代码并不难

join sep xs = foldr (\a b-> a ++ if b=="" then b else sep ++ b) "" xs
join " " ["is","there","such","a","function","?"]

joinBy sep cont = drop (length sep) $ concat $ map (\w -> sep ++ w) cont

如果有人感兴趣，还可以使用一些其他的想法实现插入和插入：

myIntersperse :: a -> [a] -> [a]
myIntersperse _ [] = []
myIntersperse e xs = init $ xs >>= (:[e])

myIntercalate :: [a] -> [[a]] -> [a]
myIntercalate e xs = concat $ myIntersperse e xs

xs >>= f等同于concat (map f xs)。

如果您想编写自己的intercalate和版本intersperse：

intercalate :: [a] -> [[a]] -> [a]
intercalate s [] = []
intercalate s [x] = x
intercalate s (x:xs) = x ++ s ++ (intercalate s xs)

intersperse :: a -> [a] -> [a]
intersperse s [] = []
intersperse s [x] = [x]
intersperse s (x:xs) = x : s : (intersperse s xs)