Haskell:没有因使用 `==' 而导致 (Eq a) 的实例

Question

isPalindrome :: [a] -> Bool
isPalindrome xs = case xs of 
                        [] -> True 
                        [x] -> True
                        a -> (last a) == (head a) && (isPalindrome (drop 1 (take (length a - 1) a)))

main = do
    print (show (isPalindrome "blaho"))

结果是

No instance for (Eq a)
  arising from a use of `=='
In the first argument of `(&&)', namely `(last a) == (head a)'
In the expression:
  (last a) == (head a)
  && (isPalindrome (drop 1 (take (length a - 1) a)))
In a case alternative:
    a -> (last a) == (head a)
         && (isPalindrome (drop 1 (take (length a - 1) a)))

为什么会出现这个错误？

Answer 1

您正在比较 a 类型的两个项目使用 == .这意味着 a不能只是任何类型 - 它必须是 Eq 的实例, 作为 == 的类型是 (==) :: Eq a => a -> a -> Bool .

您可以通过添加 Eq 来解决此问题对 a 的约束到你的函数的类型签名:

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

顺便说一句，使用 reverse 实现这个功能有一个更简单的方法。 .