从普通列表创建索引链表

Question

我目前正在玩索引链表。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？

return 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

游乐场的完整源代码is available here。

Answer 1

是的，只是使用存在主义。这将列表的长度和列表本身包装成一对，在类型中不显示它的长度。

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

这表示 SomeLList a 由 SomeLList @(n :: Nat) (_ :: LList n a) 形式的项组成。这种类型实际上等同于 []（除了一个额外的底部和没有无穷大）

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

你可以通过匹配得到类型对：

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表

create Index Linked List from a plain List

haskell

dependent-type

type-level-computation