haskell - 是否可以使用迭代增量对键入的 Church 数字实现加法？

Question

我找不到将加法定义为重复递增的方法，尽管这在无类型语言中是可能的。这是我的代码:

{-# LANGUAGE RankNTypes #-}
type Church = forall a . (a -> a) -> (a -> a)

zero :: Church
zero = \f -> id

inc :: Church -> Church
inc n = \f -> f . n f

-- This version of addition works
add1 :: Church -> Church -> Church
add1 n m = \f -> n f . m f

-- This version gives me a compilation error
add2 :: Church -> Church -> Church
add2 n m = n inc m

我得到的编译错误 add2是

Couldn't match type `forall a1. (a1 -> a1) -> a1 -> a1'
                  with `(a -> a) -> a -> a'
    Expected type: ((a -> a) -> a -> a) -> (a -> a) -> a -> a
      Actual type: Church -> (a -> a) -> a -> a
    In the first argument of `n', namely `inc'
    In the expression: n inc m
    In an equation for `add2': add2 n m = n inc m

为什么这是一个错误？不是 Church的同义词 ((a->a) -> a -> a) ?

Answer 1

无论我添加了什么额外的类型注释，我都无法输入您的代码，尽管我可能不够聪明。 (我也尝试添加 ImpredicativeTypes 。)
我认为这里的问题是在定义中

type Church = forall a. (a -> a) -> (a -> a)

a只能使用 rank-0 类型(即内部没有 foralls)实例化，即 Church本身不是。所以你不能把这样定义的 Church 数字应用到你的 inc 上。 .

但是，有一个相对简单的解决方法，它稍微有点冗长，但其他方面都可以很好地工作:make Church转换为新类型而不是类型，以便从外部将其视为单态。以下所有工作:

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

zero :: Church
zero = Church (\f -> id)

inc :: Church -> Church
inc n = Church (\f -> f . runChurch n f)

add2 :: Church -> Church -> Church
add2 n = runChurch n inc