Question

我在ghc 7.6中乱搞了数据线，并且它的运行方式并不像我认为的那样。

{-# LANGUAGE DataKinds, KindSignatures, TypeOperators #-}
import GHC.TypeLits

data Array (i :: Nat) a = Array {
  num :: Int,
  elems :: [a]
} deriving (Eq, Show)

arr10 :: Array 10 Int
arr10 = arrn 10

arr20 :: Array 20 Int
arr20 = arrn 20

arrn :: Int -> Array a Int
arrn n = Array n (replicate n 0)

arrconcat :: Array a e -> Array b e -> Array (a+b) e
arrconcat (Array a as) (Array b bs) = Array (a+b) (as ++ bs)

在ghci：

*Main> arr10 
Array {num = 10, elems = [0,0,0,0,0,0,0,0,0,0]}

*Main> arr10 == arr10
True

*Main> arr20
Array {num = 20, elems = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]}

*Main> arr10 `arrconcat` arr20
Array {num = 30, elems = [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0]}

*Main> :t arr10 `arrconcat` arr20
arr10 `arrconcat` arr20 :: Array (10 + 20) Int

*Main> :t arr10 `arrconcat` arr10 == arr20
<interactive>:1:1:
    Couldn't match type `10 + 10' with `20'
    Expected type: Array 20 Int
      Actual type: Array (10 + 10) Int
    In the first argument of `(==)', namely `arr10 `arrconcat` arr10'
    In the expression: arr10 `arrconcat` arr10 == arr20

有没有办法做我正在尝试使用这种类型级数字或者计划最终工作？

Answer 1

如果您定义类型级别的Peano数字，类型系列和数据种类实际上足以实现此功能：

{-# LANGUAGE DataKinds, TypeFamilies #-}

data Array (i :: Nat) a = Array {
  num :: Int,
  elems :: [a]
} deriving (Eq, Show)

arrn :: Int -> Array a Int
arrn n = Array n (replicate n 0)

data Nat = Zero | Succ Nat

type family Add (n :: Nat) (m :: Nat) :: Nat
type instance Add Zero m = m
type instance Add (Succ n) m = Succ (Add n m)

type Ten = Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ Zero)))))))))
type Twenty = Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ (Succ Zero)))))))))))))))))))

arr10 :: Array Ten Int
arr10 = arrn 10

arr20 :: Array Twenty Int
arr20 = arrn 20

arrconcat :: Array a e -> Array b e -> Array (Add a b) e
arrconcat (Array a as) (Array b bs) = Array (a+b) (as ++ bs)

这可以按照您的预期运作：

*Main> :t arr10 `arrconcat` arr10 == arr20
arr10 `arrconcat` arr10 == arr20 :: Bool
*Main> arr10 `arrconcat` arr10 == arr20
True

不幸的是，TypeLits目前有点未煮熟。但正如Nathan Howell已经评论过的那样，他们正在努力，并且应该在GHC 7.8中做得更好。它会很棒！

使用kindignatures和typeoperators进行类型级算术？

1 个答案: