Question

请为以下错误提供帮助。基本上对于给定的项目列表，我想创建一个函数来检查项目是否在该列表中。为了做到这一点，我将列表转换为一个集合，然后围绕该集合创建一个闭包。

toAccept :: Ord a => [a] -> (a -> Bool)
toAccept xs =
    let xset = Set.fromList xs in
    let
    -- accept:: Ord a => a -> Bool
    accept x = Set.member xset x in 
    accept

我得到的是

Could not deduce (a ~ Set.Set (Set.Set a))
from the context (Ord a)
  bound by the type signature for
             toAccept :: Ord a => [a] -> a -> Bool
  at Tokens.hs:6:13-39
  `a' is a rigid type variable bound by
      the type signature for toAccept :: Ord a => [a] -> a -> Bool
      at Tokens.hs:6:13
Expected type: a -> Bool
  Actual type: Set.Set (Set.Set a) -> Bool
In the expression: accept
In the expression: let accept x = Set.member xset x in accept
In the expression:
  let xset = Set.fromList xs in
  let accept x = Set.member xset x in accept

我做错了什么？

Answer 1

你的论点已经失败

let accept x = Set.member x xset in accept

作为一项规则，在haskell函数中，数据结构是最后一个，因为这使得它适合无点样式。有关详情，请参阅here。

可能更好的写作方式是

toAccept' :: Ord a => [a] -> a -> Bool
toAccept' = flip Set.member . Set.fromList

这里我们利用了Haskell的currying功能，因为[a] -> a -> Bool与[a] -> (a -> Bool)相同。我们实际上可以编写我们的函数，好像它接收两个参数并且不接受一个并返回一个函数，因为在Haskell中，这些选项是相同的。

无法演绎（a~Set.Set（Set.Set a））

1 个答案: