haskell - `++` 的 `foldr` 实现是如何工作的？

Question

我正在通过 Channel 9 Lectures 的 haskell 编程来学习 Haskell。在第 7 章中，教授使用 foldr 实现了自己的 ++ 运算符

(++ ys) = foldr (:) ys

他的推理是

  xs ++ ys = foldr (:) ys xs
= (++) ys xs = foldr (:) ys xs
= (++ ys) = foldr (:) ys

这是黑板

我很困惑为什么xs++ ys的功能与(++) ys xs相同。我让 xs = [1,2,3] 和 ys = [4,5,6] 并通过 runhaskell 运行它们，它们只是产生不同的答案:

> main = print $ [1,2,3] ++ [4,5,6]
[1,2,3,4,5,6]
> main = print $ foldr (:) [4,5,6] [1,2,3]
[1,2,3,4,5,6]
> main = print $ (++) [4,5,6] [1,2,3] 
[4,5,6,1,2,3]

那么教授的推理到底是什么意思呢？

顺便说一句，我正在尝试自己进行如下诱导:

  xs ++ ys = foldr (:) ys xs
= (++) xs ys = foldr(:) ys xs

等式右侧的

foldr(:) ys xs 表示此 Foldable 中的每个 y ys，将 : 应用于 y 和 xs，这实际上是在左侧所做的。我的推理正确吗？

Answer 1

IMO，您的推理是正确的 - 视频中有错误:

xs ++ ys = ...
= (++) ys xs = ...

应该是

xs ++ ys = ...
= (++) xs ys = ...

如果你看看video的评论，你会注意到 Tom Lokhorst pointed out this error (among several others) .