haskell - 如何通过可能的输出查找列表中的最大元素

Question

此代码有效

max_elem :: (Ord a) => [a] -> a 
max_elem [x] = x 
max_elem [] = error "No elements" 
max_elem (x:xs)  
    |x > max_elem xs = x 
    |otherwise = max_elem xs

我想要它，所以如果它们没有元素并且只返回 x 作为最大元素，那么它会返回 Nothing我尝试了以下方法

max_elem :: (Ord a) => [a] -> Maybe a 
max_elem [x] = Just x 
max_elem [] = Nothing
max_elem (x:xs) 
    |x >  max_elem xs = Just x
    |otherwise = max_elem xs 

我收到以下错误。请提出解决此问题的建议。

 • Couldn't match expected type ‘a’ with actual type ‘Maybe a’
      ‘a’ is a rigid type variable bound by
        the type signature for:
          max_elem :: forall a. Ord a => [a] -> Maybe a
        at <interactive>:208:13
    • In the second argument of ‘(>)’, namely ‘max_elem xs’
      In the expression: x > max_elem xs
      In a stmt of a pattern guard for
                     an equation for ‘max_elem’:
        x > max_elem xs
    • Relevant bindings include
        xs :: [a] (bound at <interactive>:211:13)
        x :: a (bound at <interactive>:211:11)
        max_elem :: [a] -> Maybe a (bound at <interactive>:209:1)

Answer 1

由于以下行，您会收到错误:x > max_elem xs。 max_elem xs 的类型为 Maybe a，其中 a 是列表的元素。它的类型为a。您无法比较不同类型的值。 a 和 Maybe a 是不同的类型。请参阅 Haskell 等式表:

• https://htmlpreview.github.io/?https://github.com/quchen/articles/blob/master/haskell-equality-table.html

将 == 运算符替换为 >，您将得到相同的表。

您可以通过将 x > max_elem xs 替换为 Just x > max_elem xs 来解决代码中的问题。这对你来说有意义吗？

如您所见，Maybe a 数据类型具有 Ord a => Ord (Maybe a) 实例，这实际上非常方便!因此，您可以使用此 Ord 实例以更简洁的方式实现您的函数:

max_elem :: Ord a => [a] -> Maybe a 
max_elem = foldr max Nothing . map Just

不过，如果您关心性能，这可能不是最有效的解决方案。