Well, I couldn't resist the puzzle. Here are solutions using foldr and
unfoldr. Don't know if they are cunning or not, but they were kind of fun.
import Data.List
splitByElem e xs =
unfoldr f xs
where f s =
case break (e ==) s of
("",_) -> Nothing
(a,b) -> Just (a, drop 1 b)
splitByElem1 e xs =
foldr f [[]] xs
where f a b = if a == e then [] : b else (a : head b) : (tail b)
J. Garrett Morris wrote:
There is at least one cunning rewriting with foldl, I think, but I
think this version is clearer.
/g
On 6/12/06, Sara Kenedy <[EMAIL PROTECTED]> wrote:
Hi all,
I want to write a function to separate a string into a list of strings
separated by commas.
Example:
separate :: String -> [String]
separate "Haskell, Haskell, and Haskell" = ["Haskell", "Haskell",
"and Haskell"]
If anyone has some ideas, please share with me. Thanks.
S.
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