Haskell错误：变量不在范围内

Question

我对 Haskell 有点陌生，但我已经解决了这个问题几个小时，但没有运气。

我正在尝试实现类似于过滤器的东西，除了一个谓词和列表被传递给一个函数，它返回一个由两个列表组成的元组，一个被谓词过滤，一个不被谓词过滤。

divideList :: (a -> Bool) -> [a] -> ([a],[a])
divideList p xs = (f, nf) where 
f = doFilter p xs
nf = doFilter (not . p) xs

doFilter :: (a -> Bool) -> [a] -> [a]
doFilter _ [] = []
doFilter p (x:xs) = [x | p x] ++ doFilter p xs

第二个函数doFilter 可以正常工作。它将过滤器应用于其列表并吐出适当的列表。 （即如果我只使用doFilter (>3) [1,2,3,4,5,6] 它将正常工作）

我的问题是第一个功能。当我使用divideList (>3) [1,2,3,4,5,6] 时，我得到了一些Variable not in scope 错误。错误如下：

AddListOperations.hs:20:23: error:
    Variable not in scope: p :: a -> Bool

AddListOperations.hs:20:25: error: Variable not in scope: xs :: [a]

AddListOperations.hs:21:31: error:
    Variable not in scope: p :: a -> Bool

AddListOperations.hs:21:34: error: Variable not in scope: xs :: [a]

就像我说的那样，我只是在玩 Haskell 几天，所以如果我遗漏了任何重要信息，请告诉我。

Answer 1

缩进f 和nf：

divideList :: (a -> Bool) -> [a] -> ([a],[a])
divideList p xs = (f, nf) where
  f  = doFilter p xs
  nf = doFilter (not . p) xs

毕竟，你的where 块会停在哪里？

顺便说一下，divideList 是来自Data.List 的partition。

Answer 2

感谢 Alec，我发现我需要做的就是缩进 where 下面的语句：

divideList :: (a -> Bool) -> [a] -> ([a],[a])
divideList p xs = (f, nf) where 
    f = doFilter p xs
    nf = doFilter (not . p) xs