haskell - (如何)在 `dependent-map` GADT 中可能具有多态值？

Question

任何人都知道如何/是否可以扩展 this code 中的 Foo GADT| :

{-# language GADTs #-}
{-# language DeriveGeneric #-}
{-# language DeriveAnyClass #-}
{-# language TemplateHaskell #-}
{-# language StandaloneDeriving #-}

import Prelude                  (Int, String, print, ($))
import Data.GADT.Show           ()
import Data.GADT.Compare        ()
import Data.Dependent.Map       (DMap, fromList, (!))
import Data.Dependent.Sum       ((==>))
import Data.GADT.Compare.TH     (deriveGEq, deriveGCompare)
import Data.Functor.Identity    (Identity)

data Foo a where
  AnInt   :: Foo Int
  AString :: Foo String

deriveGEq      ''Foo
deriveGCompare ''Foo

dmap1 :: DMap Foo Identity
dmap1 = fromList [AnInt ==> 1, AString ==> "bar"]

main = do
  print $ dmap1 ! AnInt
  print $ dmap1 ! AString

  -- Prints:
  -- Identity 1
  -- Identity "bar"

与

  ANum :: Num n => Foo n

(或类似的东西)以允许 dependent-map 中的多态值？

当我尝试时，我得到了这样的类型错误:

exp-dep-map.hs:20:1: error:
    • Couldn't match type ‘a’ with ‘b’
      ‘a’ is a rigid type variable bound by
        the type signature for:
          geq :: forall a b. Foo a -> Foo b -> Maybe (a := b)
        at exp-dep-map.hs:20:1-20
      ‘b’ is a rigid type variable bound by
        the type signature for:
          geq :: forall a b. Foo a -> Foo b -> Maybe (a := b)
        at exp-dep-map.hs:20:1-20
      Expected type: Maybe (a := b)
        Actual type: Maybe (a :~: a)
    • In a stmt of a 'do' block: return Refl
      In the expression: do return Refl
      In an equation for ‘geq’: geq ANum ANum = do return Refl
    • Relevant bindings include
        geq :: Foo a -> Foo b -> Maybe (a := b)
          (bound at exp-dep-map.hs:20:1)
   |
20 | deriveGEq      ''Foo
   | ^^^^^^^^^^^^^^^^^^^^

---

编辑:我继续与 echatav 和 isovector(GitHub 用户名)一起研究这个问题，我们能够进一步解决这个问题，我们还发现定义 GEq 和 手动 GCompare 实例有帮助。所以谢谢你，@rampion，你的回答也证实了我们的发现。

尽管为大型记录类型手动定义这些并不理想。我想知道 TemplateHaskell 生成器(deriveGCompare、deriveGEq)是否{需要、可能}更新以处理多态性。

此外，我发现对于我当前的用例，我正在寻找的 pol'ism 实际上更接近

data Foo n a where
  AnInt :: Foo n Int
  AString :: Foo n String
  ANum :: (Num n, Typeable n, Show n) => Foo n n

手动定义的实例在这里也适用，但又不太理想。

instance GEq (Foo n) where
  geq AnInt AnInt = return Refl
  geq AString AString = return Refl
  geq ANum ANum = return Refl
  geq _ _ = Nothing

instance GCompare (Foo n) where
  gcompare AnInt AnInt = GEQ
  gcompare AnInt _ = GLT
  gcompare _ AnInt = GGT
  gcompare AString AString = GEQ
  gcompare AString _ = GLT
  gcompare _ AString = GGT
  gcompare (ANum :: Foo n a) (ANum :: Foo n b) = (eqT @a @b) & \case
    Just (Refl :: a :~: b) -> GEQ
    Nothing   -> error "This shouldn't happen"
  gcompare ANum _ = GLT
  gcompare _ ANum = GGT

尝试使用 TH，(例如 deriveGEq ''Foo 或 deriveGEq ''(Foo n))我遇到了一些问题。

exp-dep-map.hs:39:1: error:
    • Expecting one more argument to ‘Foo’
      Expected kind ‘* -> *’, but ‘Foo’ has kind ‘* -> * -> *’
    • In the first argument of ‘GEq’, namely ‘Foo’
      In the instance declaration for ‘GEq Foo’
   |
39 | deriveGEq      ''Foo
   | ^^^^^^^^^^^^^^^^^^^^

exp-dep-map.hs:40:19: error: parse error on input ‘Foo’
   |
40 | deriveGEq      ''(Foo n)
   |                   ^^^

可能相关:https://github.com/mokus0/dependent-sum-template/pull/6

Answer 1

模板 haskell 很难看出发生了什么，所以我建议您滚动自己的 GEq 实例以更好地理解错误。

看the definition of GEq :

class GEq f where
  geq :: f a -> f b -> Maybe (a := b) 

我们对 a 或 b 没有任何进一步的限制，因此我们需要单独证明(或反驳)GADT 构造函数的类型相等性。

上面的 ANum 没有给我们。

不过，如果我们向 ANum 添加另一个约束 - Typeable

，这是可以解决的

ANum :: (Num n, Typeable n) => n -> Foo n

现在我们可以使用eqT见证类型平等

geq (ANum _) (ANum _) = eqT