haskell - 初学者haskell程序员

Question

isPalindrome::(Eq a) => [a] -> Bool
isPalindrome [] = True
isPalindrome [x] = True
isPalindrome (x1:xs:x2:[]) 
    | x1 == x2 = isPalindrome xs
    |otherwise = False


[1 of 1] Compiling Main             ( myHas.hs, interpreted )

myHas.hs:37:27:
    Couldn't match expected type `[a]' against inferred type `a1'
      `a1' is a rigid type variable bound by
           the type signature for `isPalindrome' at myHas.hs:33:18
    In the first argument of `isPalindrome', namely `xs'
    In the expression: isPalindrome xs
    In the definition of `isPalindrome':
        isPalindrome (x1 : xs : x2 : [])
                       | x1 == x2 = isPalindrome xs
                       | otherwise = False

我是初学者haskell程序员，不知道为什么会出现此错误，请帮忙？

Answer 1

你对待xs像一个列表，但是 (x1:xs:x2:[])假设它是您输入列表的一个元素。

请注意 (x1:xs:x2:[])将仅匹配具有 3 个元素的列表，并且 x1 , xs和 x2将是 a 类型的元素.

所以xs是 a 类型, 但是当你把它传递给 isPalindrome ，我们只能假设它一定是一个列表，所以类型系统调用类型 [a1] .

编码你想要的最简单的方法是:

isPalindrome::(Eq a) => [a] -> Bool
isPalindrome l = l == (reverse l)