教堂数字，刚性型和无限型

Question

我试图实现教会数字的前身函数pred，然后我引用了维基百科关于教会编码的页面。

根据它，我写了以下文章

{-# LANGUAGE ScopedTypeVariables, RankNTypes #-}
module Church where

newtype Church = Church { runChurch :: forall a. (a -> a) -> a -> a }

pred1 :: Church -> Church
pred1 (Church n) = Church (\f a -> extract (n (\g h -> h (g f)) (const a))) where
  extract k = k id

哪种类型的检查。

但是当我尝试使用pred1来更直接地实现它时

pred2 :: forall a. ((a -> a) -> a -> a) -> (a -> a) -> a -> a
pred2 n = runChurch $ pred1 (Church n)

ghc抱怨

    • Couldn't match type ‘a1’ with ‘a’
      ‘a1’ is a rigid type variable bound by
        a type expected by the context:
          forall a1. (a1 -> a1) -> a1 -> a1
        at Church.hs:11:30-37
      ‘a’ is a rigid type variable bound by
        the type signature for:
          pred2 :: forall a. ((a -> a) -> a -> a) -> (a -> a) -> a -> a
        at Church.hs:10:1-61
      Expected type: (a1 -> a1) -> a1 -> a1
        Actual type: (a -> a) -> a -> a
    • In the first argument of ‘Church’, namely ‘n’
      In the first argument of ‘pred1’, namely ‘(Church n)’
      In the second argument of ‘($)’, namely ‘pred1 (Church n)’
    • Relevant bindings include
        n :: (a -> a) -> a -> a (bound at Church.hs:11:7)
        pred2 :: ((a -> a) -> a -> a) -> (a -> a) -> a -> a
          (bound at Church.hs:11:1)
   |
11 | pred2 n = runChurch $ pred1 (Church n)
   |                                     ^

或者是lambda演算风格

pred3 :: forall a. ((a -> a) -> a -> a) -> (a -> a) -> a -> a
pred3 n f a1 = extract (n (\g h -> h (g f)) (const a1)) where
  extract k = k id

编译器说

    • Occurs check: cannot construct the infinite type:
        a ~ (a0 -> a0) -> a
    • In the first argument of ‘extract’, namely
        ‘(n (\ g h -> h (g f)) (const a1))’
      In the expression: extract (n (\ g h -> h (g f)) (const a1))
      In an equation for ‘pred3’:
          pred3 n f a1
            = extract (n (\ g h -> h (g f)) (const a1))
            where
                extract k = k id
    • Relevant bindings include
        a1 :: a (bound at Church.hs:14:11)
        f :: a -> a (bound at Church.hs:14:9)
        n :: (a -> a) -> a -> a (bound at Church.hs:14:7)
        pred3 :: ((a -> a) -> a -> a) -> (a -> a) -> a -> a
          (bound at Church.hs:14:1)
   |
14 | pred3 n f a1 = extract (n (\g h -> h (g f)) (const a1)) where
   |                         ^^^^^^^^^^^^^^^^^^^^^^^^^^

如果我不指定pred3的类型，推断的类型是

pred3 :: (((t1 -> t2) -> (t2 -> t3) -> t3) -> (b -> a1) -> (a2 -> a2) -> t4)
    -> t1 -> a1 -> t4

我找不出这两个错误，任何建议都会有帮助

Answer 1

想象一下有人试图给pred2 @Int打电话。然后，您将尝试将一个(Int -> Int) -> Int -> Int填充到一个Church中，这显然是错误的。要使pred2工作，您需要确保它的第一个参数始终是一个多态函数，这意味着pred2需要一个秩-2类型。它的定义很好，所以只需将其类型签名替换为：

pred2 :: (forall a. (a -> a) -> a -> a) -> (b -> b) -> b -> b

这同样适用于pred3。