How to Code the Function Drop Without the Prelude Module in Haskell?
I Want to Write Some Prelude Functions by Myself in Haskell to Understand It Better and to Exercise a Bit with It. I Already Coded the Take Function but I Have...
I want to write some prelude functions by myself in Haskell to understand it better and to exercise a bit with it. I already coded the TAKE function but I have some problems with the DROP function.
take1 :: Int -> [a] -> [a]
take1 _ [] = []
take1 0 _ = []
take1 n (x:xs) = x:take1 (n-1) xs
-- Input: take 5 [1,2,3,4,5,6,7]
-- Output: [1,2,3,4,5]
--This is the take Function and its working.
drop1 :: Int -> [a] -> [a]
drop1 _ [] = []
drop1 n (x:xs) = x : drop1 (n+1) xs
-- Input: drop 5 [1,2,3,4,5,6,7,8,9,10]
-- Output: [6,7,8,9,10]
-- It´s printing out the whole List and not the wanted result.
1 Answer
The idea of drop is to skip the n first items, so just discard them recursively until you reach 0:
drop:: Int -> [a] -> [a]
drop _ [] = []
drop 0 xs = xs
drop n (_:xs) = drop (n-1) xs
The important base case that you miss is drop 0 xs = xs, so if I have nothing to drop I just return what I got.
Also note that we decrement ((n-1)) not increment in the recursive call, otherwise you will never reach to the base case.
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