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where子句中的Haskell解析器错误

haskell syntax

5

Hynek -Pichi- Vychodil Paulo Suassuna · 技术社区 · 17 年前

你怎么了 rs

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困惑。错误消息是

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

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

2 回复 | 直到 14 年前

1

14

Calibre Johannes Schaub - litb 9 年前

你的缩进是错误的,我想你只能有一个 where 在那里(我可能大错特错,我不是哈斯凯尔人)。还缺少一个调用的参数 rev

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")

"helloolleh"

我现在要试着去理解它。无论如何,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

2

0

Hynek -Pichi- Vychodil Paulo Suassuna 17 年前

@利特:你可以用这种方式重写con

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