It won't compile because the type of (:) is (:) :: a -> [a] -> [a]. It takes an element on its left side. take returns a list (take :: Int -> [a] -> [a]). Instead of using the (:) operator, use the (++) operator. It appends a list to another. So, your function should be: myremove x s = (take (s-1) x) ++ (drop (s+1) x) But it won't do what you're expecting it to do because, drop 3 "doaltan" would remove doa" and give you "lo". It removes the first 3 elements. So, your drop (s+1) would remove (3+1), elements, i.e, "doal". Change it to drop s. So, your final function should be: myremove x s = (take (s-1) x) ++ (drop s x) And by the way, welcome to Haskell! On Sun, Feb 19, 2012 at 11:30 PM, bahadýr altan wrote:

Hello. I'm trying to write a code which deletes the wanted element in a list. Like this :  myremove "doaltan" 3 = "doltan". I wrote the code below but it doesn't even compile. Could you help me with fixing the code? Thanks!

myremove x s = (take  (s-1) x):(drop (s+1) x)

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-- Warm Regards, AbdulSattar Mohammed

Look at the types of (:), take, and drop. You want to concat lists, not append an element to the head of a list. I would recommend looking at the concat operator (++). Haskell is all about the types! take returns a list, as does drop. So you need some function of type [a] -> [a] -> [a], not a -> [a] -> [a].