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!
--
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explain-me-how-drop-works-in-Haskell-tf3290490.html#a9152251
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