Haskell无法统一类型实例方程

我试图用（偶数/奇数）奇偶校验类型标记规范Nat数据类型，看看我们是否能得到任何自由定理。代码如下：

{-# LANGUAGE GADTs #-}
{-# LANGUAGE TypeFamilies #-}
{-# LANGUAGE DataKinds #-}

-- Use DataKind promotion with type function for even-odd

module EvenOdd where

  data Parity = Even | Odd
  -- Parity is promoted to kind level Parity.
  -- Even & Odd to type level 'Even & 'Odd of kind Parity

  -- We define type-function opp to establish the relation that
  -- type 'Even is opposite of 'Odd, and vice-versa
  type family Opp (n :: Parity) :: Parity
  type instance Opp 'Even = 'Odd
  type instance Opp 'Odd = 'Even

 -- We tag natural number with the type of its parity
  data Nat :: Parity -> * where
     Zero :: Nat 'Even
     Succ :: Nat p -> Nat (Opp p)

  -- Now we (should) get free theorems.
  -- 1. Plus of two even numbers is even
  evenPlus :: Nat 'Even -> Nat 'Even -> Nat 'Even
  evenPlus Zero n2 = n2  -- Line 31
  evenPlus (Succ (Succ n1)) n2 = Succ (Succ (evenPlus n1 n2))

但是，GHC抛出类型错误：

Could not deduce (p1 ~ 'Even)
from the context ('Even ~ Opp p)
  bound by a pattern with constructor
             Succ :: forall (p :: Parity). Nat p -> Nat (Opp p),
           in an equation for `evenPlus'
  at even-odd.hs:31:13-26
or from (p ~ Opp p1)
  bound by a pattern with constructor
             Succ :: forall (p :: Parity). Nat p -> Nat (Opp p),
           in an equation for `evenPlus'
  at even-odd.hs:31:19-25
  `p1' is a rigid type variable bound by
       a pattern with constructor
         Succ :: forall (p :: Parity). Nat p -> Nat (Opp p),
       in an equation for `evenPlus'
       at even-odd.hs:31:19
Expected type: Nat 'Even
  Actual type: Nat p
In the first argument of `evenPlus', namely `n1'
In the first argument of `Succ', namely `(evenPlus n1 n2)'

据我所知，上述错误的要点是，当上下文有以下等式时，GHC无法推断（p1~'偶数）（（Opp（Opp p1））~'偶数）

---

为什么会发生这种情况？我的方法有什么问题吗？

我不认为GADT模式匹配优化可以这样工作。您将

Opp

作为构造函数的结果类型。所以如果你写了这样的东西

f :: Nat 'Even -> ...
f (Succ n) = ...

然后类型检查器知道

Nat（Opp t）~Nat'偶数

，因此

Opp t~偶数

。但要解决这个问题，类型检查器必须反转函数

Opp

，这要求很多

我建议您将

Nat

的定义改为：

data Nat :: Parity -> * where
  Zero :: Nat 'Even
  Succ :: Nat (Opp p) -> Nat p

这应该行得通

编辑 实际上，让我稍微扩展一下

上述建议并非没有（次要）价格。你失去了一点类型推断。例如，

succzero

的类型现在是

succzero:：Opp~'偶数=>natp

而不是

Nat'Odd

。通过显式类型注释，它解析为ok

您可以通过向

Succ

添加一个约束来改进这一点，该约束要求

Opp

是自反转的。

奇偶校验

仅有的两个元素是

偶数

和

奇数

，对于这两个元素，约束是成立的，因此它不会导致任何问题：

data Nat :: Parity -> * where
  Zero :: Nat 'Even
  Succ :: (Opp (Opp p) ~ p) => Nat (Opp p) -> Nat p

---

现在

succzero

被推断为

Nat'Odd

类型，模式匹配仍然有效。

这有效，谢谢。我发现这两种方法在生成的类型约束方面有所不同。@GowthamKaki我在我的答案中添加了更多的细节。