Haskell 无法在两个存在论中就单例库推断KnownNat
我在试验Singleton库,发现了一个我不理解的案例Haskell 无法在两个存在论中就单例库推断KnownNat,haskell,dependent-type,existential-type,Haskell,Dependent Type,Existential Type,我在试验Singleton库,发现了一个我不理解的案例 {-# LANGUAGE GADTs, StandaloneDeriving, RankNTypes, ScopedTypeVariables, FlexibleInstances, KindSignatures, DataKinds, StandaloneDeriving #-} import Data.Singletons.Prelude import Data.Singletons.TypeLits data Foo (a ::
{-# LANGUAGE GADTs, StandaloneDeriving, RankNTypes, ScopedTypeVariables,
FlexibleInstances, KindSignatures, DataKinds, StandaloneDeriving #-}
import Data.Singletons.Prelude
import Data.Singletons.TypeLits
data Foo (a :: Nat) where
Foo :: Foo a
deriving Show
data Thing where
Thing :: KnownNat a => Foo a -> Thing
deriving instance Show Thing
afoo1 :: Foo 1
afoo1 = Foo
afoo2 :: Foo 2
afoo2 = Foo
athing :: Thing
athing = Thing afoo1
foolen :: forall n. KnownNat n => Foo n -> Integer
foolen foo =
case sing of (SNat :: Sing n) -> natVal (Proxy :: Proxy n)
minfoo :: forall a b c. (Min a b ~ c, KnownNat c) => Foo a -> Foo b -> Integer
minfoo _ _ =
let c = case sing of (SNat :: Sing c) -> natVal (Proxy :: Proxy c)
in natVal (Proxy :: Proxy c)
thinglen :: Thing -> Integer
thinglen (Thing foo) = foolen foo
我可以用这个来得到至少两样东西
minthing :: Thing -> Thing -> Integer
minthing (Thing foo1) (Thing foo2) = min (foolen foo1) (foolen foo2)
但是为什么我不能这么做呢
minthing' :: Thing -> Thing -> Integer
minthing' (Thing foo1) (Thing foo2) = minfoo foo1 foo2
• Could not deduce (KnownNat
(Data.Singletons.Prelude.Ord.Case_1627967386
a
a1
(Data.Singletons.Prelude.Ord.Case_1627967254
a a1 (GHC.TypeLits.CmpNat a a1))))
你需要做一些定理证明来检查给定的
KnownNat a
和KnownNat b
你可以得到KnownNat(Min a b)
。一种可能的解决办法:
import Data.Constraint
(……)
我觉得user3237465的评论需要永久化,因为它删除了containt库依赖项,而且非常简洁
minthing' :: Thing -> Thing -> Integer
minthing' (Thing foo1) (Thing foo2) =
theorem (fooSing foo1) (fooSing foo2) $ minfoo foo1 foo2
where
fooSing :: KnownNat a => Foo a -> Sing a
fooSing = const sing
theorem :: forall a b c. (KnownNat a, KnownNat b) =>
Sing a -> Sing b -> (KnownNat (Min a b) => c) -> c
theorem sa sb c = case sCompare sa sb of
SLT -> c
SEQ -> c
SGT -> c
我不得不承认,我在这方面做得不多,但在我看来,你在
minfoo
中缺少(KnownNat a,KnownNat b)
。
minthing' :: Thing -> Thing -> Integer
minthing' (Thing foo1) (Thing foo2) =
theorem (fooSing foo1) (fooSing foo2) $ minfoo foo1 foo2
where
fooSing :: KnownNat a => Foo a -> Sing a
fooSing = const sing
theorem :: forall a b c. (KnownNat a, KnownNat b) =>
Sing a -> Sing b -> (KnownNat (Min a b) => c) -> c
theorem sa sb c = case sCompare sa sb of
SLT -> c
SEQ -> c
SGT -> c