syntax - 在 where 子句中的 Haskell 解析器错误

Question

rs 有什么问题|第一个 where 部分的定义？

palindrome :: [a] -> [a]

palindrome xs = con xs rs
    where con a b = rev (rev a []) b
        rs = rev xs                        -- here
        where rev [] rs = rs
            rev (x:xs) rs = rev xs (x:rs)

我只是在学习 Haskell，但它的语法规则让我感到困惑。错误信息是

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

pelindrome.hs:5:8: parse error on input `rs'

Answer 1

你的缩进是错误的，我认为你只能有一个 where在那里(我可能错了。我不是haskell人)。对 rev 的调用也缺少一个参数。 (一个空列表):

palindrome :: [a] -> [a]
palindrome xs = con xs rs
    where con a b = rev (rev a []) b
          rs = rev xs []                       -- here
          rev [] rs = rs
          rev (x:xs) rs = rev xs (x:rs)

main = print (palindrome "hello")

打印出来:

"helloolleh"

我现在要试着理解它。不管怎样，玩得开心!

编辑:现在对我来说很有意义。我认为这是正确的版本。有关 Haskell 缩进规则，请阅读 Haskell Indentation