Question

I’m writing a function in Haskell that recursively returns the list with the specified number of elements removed from the front of the list. I’ve gotten it to work doing this:

removefront :: Int -> [Int] -> [Int]
removefront n xs =
    if n <= 0 then xs
    else removefront (n-1) (tail xs)

This works and does exactly what I want however is there a way to do the same thing without the tail function. Thanks!

Answer 1

this function is “built in” in the sense that it is in the prelude and called drop

*Main> :t drop
drop :: Int -> [a] -> [a]
*Main> drop 3 [1,2,3,4,5,6,7]
[4,5,6,7]

now, I will assume that is not the answer you were looking for. You can easily modify your function to not use tail. The trick is to use pattern matching.

removefront :: Int -> [Int] -> [Int]
removefront n (x:xs) = if n <= 0 then (x:xs) else removefront (n-1) xs

three notes

1. Most Haskeller would not use if then else for such a function, preferring guards

   removefront n (x:xs) 
      | n <= 0    = (x:xs)
      | otherwise = removefront (n-1) xs
   

2. The type of removefront can be much more general

   removefront :: Int -> [a] -> [a]
   

   actually it could be all the way to

   removefront :: (Num i, Ord i) => i -> [a] -> [a]
   

   but that is getting excessive

3. you should consider what happens when you hand your function the empty list–what do you want it to do?