haskell - 在haskell中用foldr分割

Question

我需要使用foldr编写一个函数split

split :: Eq a ⇒ a → [a] → [[ a ]]

示例:

split '/' ”hello/my/friends” ----> [”hello”,”my”,”friends”]

这是我尝试过的:

split :: Eq a ⇒ a → [a] → [[ a ]]
split str delim = let (start, end) = break (== delim) str
                in start : if null end then [] else groupBy (tail end) delim 

Answer 1

这样的事情应该有效:

foldr (\c (x:xs) ->
    if c == '/'
    then "":x:xs
    else (c:x):xs
  ) [""] "hello/my/friends"

广义:

split on = foldr (\c (x:xs) ->
    if c == on
    then []:x:xs
    else (c:x):xs
  ) [[]]