Question

My task is to re-implement this function

divn :: Integer -> [Integer] -> [Integer]
divn _ [] = []
divn n (x:xs) | mod x n == 0 = x : divn n xs
              | otherwise = divn n xs

using ‘foldr’.
What I did:

divn' _ [] = []
divn' n (x:xs) = foldr (\x -> if (mod x n == 0) (x:) ([]++)) [] xs

I thought this would work. Actually it doesn’t even compile, but says: “Parse error on input “)”.

As I didn’t find any errors, I decided to re-write if as if’ an now its working…

if' True x _ = x
if' False _ x = x

divn' _ [] = []
divn' n (x:xs) = foldr (\x -> if' (mod x n == 0) (x:) ([]++)) [] xs

Does anyone know where’s the error?

Thanks!

Answer 1

if needs a then and an else in Haskell,

(\x -> if (mod x n == 0) (x:) ([]++))

should be

(\x -> if (mod x n == 0) then (x:) else id)