On Oct 21, 2009, at 3:16 AM, satorisanitarium wrote:
I just started learning haskell and I just came to my first wtf moment.
I'm trying to do something like this:
calling: foo 3 [1,2,3,2,4,1,6,3,6,2,3,5,2,5,2,1,6,4]
returns: [[1,2,3],[2,4,1,6,3],[2,3]]
but i have no idea how to put something into a sublist.
From the way you talk about the problem, I suspect that you are trying to understand in a rather complicated way. Here's how I approach your example: foo item list = if there is an occurrence of item in list then a new list whose first element is everything up to and including that occurrence and whose remaining elements are found by applying foo item to everything after that occurrence else an empty list We can do this in just a couple of lines using List.elemIndex and splitAt, but let's do it in an elementary way. We'll write a helper function that either returns (before-and-including-item, after-item) or nothing. find_and_split :: Eq t => t -> [t] -> Maybe ([t], [t]) find_and_split item list = loop list [] where loop [] _ = Nothing loop (x:xs) before | x == item = Just (reverse (x:before), xs) loop (x:xs) before = loop xs (x:before) foo :: Eq t => t -> [t] -> [[t]] foo item list = case find_and_split item list of Nothing -> [] Just (before, after) -> before : foo item after The answer I get is *Main> foo 3 [1,2,3,2,4,1,6,3,6,2,3,5,2,5,2,1,6,4] [[1,2,3],[2,4,1,6,3],[6,2,3]] ^ which differs from yours at the marked place, but which I think is right. List.elemIndex tells you where an element occurs. Just 0 is the first position. splitAt splits a list into two pieces, the size argument tells you the length of the first piece. So foo item list = case List.elemIndex item list of Nothing -> [] Just pos -> before : foo item after where (before, after) = splitAt (pos+1) list This gives the same answer as the previous version. In neither case do we do ANYTHING to "put" items into a "sublist", we just construct a list whose elements happen to be lists. It's no different from calculating a list of numbers.