Hey, you're almost there: drop :: Integer -> [a] -> [a] drop 0 xs = xs drop n (x:xs) = drop (n-1) xs Your version fails when trying to do drop 10 [1..10]. My version fails when trying to do drop 10 [1..9], so you might want to try to see if you can come up with a solution for that! Good luck, -chris On 25 Feb, 2007, at 18:43 , iliali16 wrote:
Hi I am trying to implement the function drop in haskell the thing is that I I have been trying for some time and I came up with this code where I am trying to do recursion:
drop :: Integer -> [Integer] -> [Integer] drop 0 (x:xs) = (x:xs) drop n (x:xs) |n < lList (x:xs) = dropN (n-1) xs : |otherwise = []
So I want to understand how would this work and what exacttly should I put as an answer on line 4 couse that is where I am lost. I know I might got the base case wrong as well but I don't know what to think for it. I have done the lList as a function before writing this one. Thanks to those who can help me understand this. Thanks alot in advance! Have a nice day! -- View this message in context: http://www.nabble.com/Hi-can-u- explain-me-how-drop-works-in-Haskell-tf3290490.html#a9152251 Sent from the Haskell - Haskell-Cafe mailing list archive at Nabble.com.
_______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe