Question

I have written the code below to test whether a list is a Palindrome or not. To my surprise its not compiling with the error “No instance for (Eq a) arising from a use of == ….”. My assumption is that the compiler does not know that leftHalf and rightHalf are lists.

    isPalindrome :: [a] -> Bool
    isPalindrome [] = False
    isPalindrome xs = if (leftHalf == reverse rightHalf)
              then True
              else False
     where leftHalf = take center xs
           rightHalf = drop center xs
           center = if  even (length xs)
                       then (length xs) `div` 2
                   else ((length xs) - 1) `div` 2 

1) How do I tell the compiler that leftHalf and rightHalf are lists?
2) How would I use pattern matching or other haskell language features to solve this?

EDIT: Thank you all for your input. Special mention to Matt Fenwick for the documentation link and Duri for the elegant tip. I will write the final solutions below just in case

     isPalindrome' :: (Eq a) => [a] -> Bool
     isPalindrome' [] = False
     isPalindrome' xs = if p then True else False
                   where p = leftHalf == rightHalf
                         leftHalf = take c xs
                         rightHalf = take c (reverse xs)
                         c = div l 2
                         l = length xs

isPalindrome' can be improved like Demi pointed out

      isPalindrome'' :: (Eq a) => [a] -> Bool
      isPalindrome'' [] = False
      isPalindrome'' xs = if (reverse xs) == xs then True else False

Answer 1

Check out the Eq typeclass:

ghci> :i (==)
class Eq a where
  (==) :: a -> a -> Bool
  ...
    -- Defined in GHC.Classes
infix 4 ==

What you need is a type constraint on isPalindrome.

Also, this code

if (leftHalf == reverse rightHalf)
              then True
              else False

is unnecessarily long.