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.}

Answer 2

hammar explanation is correct.

Another simple example:

nosPrimeiros :: a -> [(a,b)] -> Bool
nosPrimeiros e [] = False
nosPrimeiros e ((x,y):rl) = if (e==x)   then True
                                        else nosPrimeiros e rl

The (e==x) will fail for this function signature. You need to replace:

nosPrimeiros :: a -> [(a,b)] -> Bool

adding an Eq instance for a

nosPrimeiros :: Eq => a -> [(a,b)] -> Bool

This instantiation is saying that now, a has a type that can be comparable with and (e==x) will not fail.