haskell 在Hashkell绑定中使用where或let时出现语法错误

hts6caw3 于 2022-12-23 发布在其他

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我最近才接触到Haskell世界。但是，仍然在语法上挣扎。后面的两个方法都有语法错误。问题可能是什么？版本1：

rev' :: [a] -> [a]
rev' [] = []
rev' [x] = [x]
rev' (x:xs) = let revHelper acc l = 
                    | acc [] = []
                    | revHelper acc (x':l') = revHelper (x':acc) l'
              in revHelper [] (x:xs)

第二版：

rev' :: [a] -> [a]
rev' [] = []
rev' [x] = [x]
rev' (x:xs) = revHelper [] (x:xs)
    where revHelper acc l = 
            | acc [] = []
            | revHelper acc (x':l') = revHelper (x':acc) l'

两者都有同样的问题：

util.hs:47:21: error: parse error on input ‘|’
   |
47 |                     | acc [] = []
   |

有人能帮帮我吗？
我正在尝试实现一个尾部递归反向函数。

haskell

来源：https://stackoverflow.com/questions/74844603/syntax-error-using-where-or-let-in-binding-in-hashkell

1条答案

按热度按时间

kcwpcxri1#

您使用防护的方式不正确。您可能希望使用以下方式实现此问题：

rev' :: [a] -> [a]
rev' [] = []
rev' [x] = [x]
rev' (x:xs) = let revHelper acc [] = []
                  revHelper acc (x':l') = revHelper (x':acc) l'
              in revHelper [] (x:xs)

但这并不完全正确：您应该返回累加器，因此：

rev' :: [a] -> [a]
rev' [] = []
rev' [x] = [x]
rev' (x:xs) = let revHelper acc [] = acc
                  revHelper acc (x':l') = revHelper (x':acc) l'
              in revHelper [] (x:xs)

此外，这里没有必要定义rev'的基本情况，您可以将其实现为：

rev' :: [a] -> [a]
rev' = go []
    where go acc [] = acc
          go acc (x:xs) = go (x:acc) xs

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haskell 在Hashkell绑定中使用where或let时出现语法错误

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