有没有任何机制来强制Haskell (除了unsafeCoerce,我希望这是可行的)约束?
{-# LANGUAGE AllowAmbiguousTypes #-}
{-# LANGUAGE ConstraintKinds #-}
{-# LANGUAGE DerivingVia #-}
{-# LANGUAGE GADTs #-}
{-# LANGUAGE RankNTypes #-}
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE StandaloneDeriving #-}
{-# LANGUAGE StandaloneKindSignatures #-}
{-# LANGUAGE TypeApplications #-}
module CatAdjonctionsSOQuestion where
import Data.Proxy
import Data.Tagged
import Unsafe.Coerce
newtype K a ph = K {unK :: a} -- I would want c a => c ((K a) i) for any c :: Constraints
-- I could do any possible instance by hand
deriving via a instance Semigroup a => Semigroup ((K a) i)
-- I want them all
-- deriving via a instance c ((K a) i) -- Instance head is not headed by a class: c (K a i)
data Exists c where
Exists :: c a => a -> Exists c
data ExistsKai c i where
ExistsKai :: c ((K a) i) => Proxy a -> ExistsKai c i
ok :: forall x c i. (forall x. (forall a. c a => a -> x) -> x) -> (forall a. c ((K a) i) => Tagged a x) -> x
ok s k =
let e = (s Exists :: Exists c)
in let f = unsafeCoerce e :: ExistsKai c i
in case f of (ExistsKai (Proxy :: Proxy a)) -> unTagged (k @a)发布于 2021-05-16 16:45:38
只要稍微修改一下,你就会要求进行亲切的检查。
newtype K a ph = K {unK :: a}
-- I would want c a => c ((K a) i)
-- for any c :: Type -> Constraint不管是现在还是将来,你绝对不能得到它,因为它是无效的。考虑一下
(~) Bool :: Type -> Constraint现在(~) Bool Bool保持不变,但您永远无法实现(~) Bool (K Bool i)。
如果没有平等约束呢?我也可以这样做,用莱布尼兹等式:
class Bar a where
isBool :: f a -> f Bool
instance Bar Bool where
isBool = id但是,没有办法编写instance Bar (K Bool i),因为它的isBool没有触底。
https://stackoverflow.com/questions/67557365
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