Haskell: No instance for (Eq a) arising from a use of `=='

Question

isPalindrome :: [a] -> Bool
isPalindrome xs = case xs of 
                        [] -> True 
                        [x] -> True
                        a          


        

Answer 1

You're comparing two items of type a using ==. That means a can't just be any type - it has to be an instance of Eq, as the type of == is (==) :: Eq a => a -> a -> Bool.

You can fix this by adding an Eq constraint on a to the type signature of your function:

isPalindrome :: Eq a => [a] -> Bool

_{By the way, there is a much simpler way to implement this function using reverse.}