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尽管已明确注释,但 Haskell 无法推断类型(或类型级别的 Nat)等式?

转载 作者:行者123 更新时间:2023-12-04 19:27:38 29 4
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我试图用 Haskell 实现一个 Braun 树,定义如下:

{-# LANGUAGE GADTs #-}
{-# LANGUAGE DataKinds #-}
{-# LANGUAGE PolyKinds #-}
{-# LANGUAGE KindSignatures #-}
{-# LANGUAGE TypeOperators #-}
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE FlexibleInstances #-}
{-# LANGUAGE TypeApplications #-}
{-# LANGUAGE ScopedTypeVariables #-}

data BraunTree (n :: Nat) a where
Empty :: BraunTree 0 a
Fork :: a -> BraunTree n a ->
BraunTree m a ->
Either (n :~: m) (n :~: (m + 1)) ->
BraunTree (n + m + 1) a

现在,我正在尝试尝试如何“类型安全地”将内容插入到这棵树中。
insert :: a -> BraunTree (n :: Nat) a -> BraunTree (n + 1 :: Nat) a
insert x Empty = Fork x Empty Empty (Left Refl)
insert x (Fork y (t1 :: BraunTree p a) (t2 :: BraunTree q a) (Left (Refl :: p :~: q))) = Fork x (t1' :: BraunTree (p + 1) a) (t2 :: BraunTree q a) (Right (sucCong Refl :: (p + 1) :~: (q + 1)))
where
t1' :: BraunTree (p + 1) a
t1' = insert x t1

sucCong作为
sucCong :: ((p :: Nat) :~: (q :: Nat)) -> (p + 1 :: Nat) :~: (q + 1 :: Nat)
sucCong Refl = Refl

现在,虽然 insert 的第一个子句编译正常,第二行抛出一个令人困惑的错误。
/home/agnishom/test/typeExp/braun.hs:31:90: error:
• Could not deduce: (((n1 + 1) + n1) + 1) ~ (n + 1)
from the context: n ~ ((n1 + m) + 1)
bound by a pattern with constructor:
Fork :: forall a (n :: Nat) (m :: Nat).
a
-> BraunTree n a
-> BraunTree m a
-> Either (n :~: m) (n :~: (m + 1))
-> BraunTree ((n + m) + 1) a,
in an equation for ‘insert’
at /home/agnishom/test/typeExp/braun.hs:31:11-85
or from: m ~ n1
bound by a pattern with constructor:
Refl :: forall k (a :: k). a :~: a,
in an equation for ‘insert’
at /home/agnishom/test/typeExp/braun.hs:31:69-72
Expected type: BraunTree (n + 1) a
Actual type: BraunTree (((n1 + 1) + m) + 1) a
NB: ‘+’ is a type function, and may not be injective
• In the expression:
Fork
x
(t1' :: BraunTree (p + 1) a)
(t2 :: BraunTree q a)
(Right (sucCong Refl :: (p + 1) :~: (q + 1)))
In an equation for ‘insert’:
insert
x
(Fork y
(t1 :: BraunTree p a)
(t2 :: BraunTree q a)
(Left (Refl :: p :~: q)))
= Fork
x
(t1' :: BraunTree (p + 1) a)
(t2 :: BraunTree q a)
(Right (sucCong Refl :: (p + 1) :~: (q + 1)))
where
t1' :: BraunTree (p + 1) a
t1' = insert x (t1 :: BraunTree p a)
• Relevant bindings include
t1' :: BraunTree (n1 + 1) a
(bound at /home/agnishom/test/typeExp/braun.hs:34:9)
t1 :: BraunTree n1 a
(bound at /home/agnishom/test/typeExp/braun.hs:31:19)
insert :: a -> BraunTree n a -> BraunTree (n + 1) a
(bound at /home/agnishom/test/typeExp/braun.hs:29:1)

我不确定我在这里做错了什么。另外,为什么haskell 认为 t1 :: BraunTree n1 a (在错误消息中),即使我已经注释了 t1 :: BraunTree p a ?

帮助解释此错误消息将非常有帮助

最佳答案

您可以尝试使用此编译器插件来推断 Nat 的类型相等性s 自动为您:

  • https://github.com/clash-lang/ghc-typelits-natnormalise
  • 关于尽管已明确注释,但 Haskell 无法推断类型(或类型级别的 Nat)等式?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51290745/

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