Question

我无法找到一种方法来将加法定义为重复递增，尽管这可能是一种无类型语言。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

我为add2获得的编译错误是

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 ((a->a) -> a -> a)的同义词吗？

Answer 1

无论我添加哪种类型的注释，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。）我认为问题在于定义

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

a只能用rank-0类型实例化（即内部没有foralls），而Church本身不是。{1}}。因此，您无法将以这种方式定义的教会数字应用于inc。

然而，有一个相对简单的解决方法略显冗长但是使一切工作都很好：将Church变成newtype而不是类型，这样它就可以被视为来自外部的单态。以下一切都有效：

{-# 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

是否可以使用迭代递增在类型化的教会数字上实现添加？

1 个答案: