Question

我试图编写一个程序来检查列表是否是回文并返回布尔。

isPalindrome :: [a] -> Bool
isPalindrome [] = True
isPalindrome [x] = True
isPalindrome xs | (head xs) == (last xs) = isPalindrome (init(tail xs))
                | otherwise = False

我收到这样的错误消息：

problem6.hs:4:19: error:
    * No instance for (Eq a) arising from a use of `=='
      Possible fix:
        add (Eq a) to the context of
          the type signature for:
            isPalindrome :: forall a. [a] -> Bool
    * In the expression: (head xs) == (last xs)
      In a stmt of a pattern guard for
                     an equation for `isPalindrome':
        (head xs) == (last xs)
      In an equation for `isPalindrome':
          isPalindrome xs
            | (head xs) == (last xs) = isPalindrome (init (tail xs))
            | otherwise = False
  |
4 | isPalindrome xs | (head xs) == (last xs) = isPalindrome (init(tail xs))
  |                   ^^^^^^^^^^^^^^^^^^^^^^
Failed, no modules loaded.

由于我不是非常有经验，所以我不理解错误消息中的内容。因此，我看不出代码中的错误所在。感谢您的帮助。

Answer 1

问题是您需要约束多态类型unary_op。目前，编译器没有有关类型的信息，因此它甚至不知道是否为a定义了(==)（这就是a的来源。它试图为No instance for (Eq a) arising from a use of ``=='推断一个Eq的实例，但是无法。您需要帮助）。

您应该输入：

现在您要告诉它，isPalindrome :: (Eq a) => [a] -> Bool仅能得到属于isPalindrome实例的事物的列表。

指出这一点是因为您正在尝试比较两个Eq的相等性：

有关错误消息的一些信息：

(head xs) == (last xs)

在我的建议中，Possible fix: add (Eq a) to the context of the type signature for: isPalindrome :: forall a. [a] -> Bool之前的内容称为上下文，您可以在其中为类型添加约束。这里的建议是告诉您完全按照我在上面所说的做（尽管以更详细的方式）。

Answer 2

import Data.List

isPalindrome :: (Eq a) => [a] -> Bool
isPalindrome [] = True
isPalindrome [x] = True
isPalindrome xs    
                | (head xs /= head (reverse xs)) = False
                | otherwise = isPalindrome ( tail (init xs) )

Haskell中的isPalindrome函数给出错误

2 个答案: