Question

我目前正在使用索引链接列表。基本数据类型由

给出

data LList (n :: Nat) (a :: Type) where
  Nil  ::LList 0 a
  (:@) ::a -> LList n a -> LList (n + 1) a

我想知道是否可以定义从[]到LList的映射吗？

返回类型取决于运行时信息，因为列表的长度当然在编译时不可用。

fromList :: ? => [a] => LList ? a
fromList = undefined

游乐场is available here的完整源代码。

Answer 1

是的，只需使用一个存在的。这样会将列表的长度和列表本身包装为一对，但不会在类型中显示其长度。

data SomeLList a = forall n. SomeLList (LList n a)

这表示SomeLList a由SomeLList @(n :: Nat) (_ :: LList n a)形式的术语组成。实际上，这种类型等效于[]（除了多余的底数和无穷大之外）

fromList :: [a] -> SomeLList a
fromList [] = Nil
fromList (x : xs) | SomeList xs' <- fromList xs = SomeList (x :@ xs)

您可以通过匹配来获得配对中的类型：

something :: [a] -> ()
something xs
  | SomeList xs' <- fromList xs
  = -- here, xs' :: SomeList n xs, where n :: Nat is a new type (invisibly) extracted from the match
    -- currently, we don't know anything about n except its type, but we could e.g. match on xs', which is a GADT and could tell us about n
    ()

从普通列表创建索引链接列表

1 个答案: