Haskell 错误 : Variable not in scope

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，除了谓词和列表被传递给一个函数，它返回两个列表的元组，一个由谓词过滤，一个不是。

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工作正常。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]我得到一个 Variable not in scope的号码错误。下面列出了错误:

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，我只在 Haskell 上玩了几天，所以如果我遗漏了任何重要信息，请告诉我。

Answer 1

两者都缩进 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，您的 where 在哪里？挡住？

顺便说一句， divideList是 partition来自 Data.List .