`foldr` 的这个 `++` 实现是如何工作的？

Question

我正在第 9 频道的讲座中学习 Haskell 编程 haskell。在第 7 章中，教授通过

使用 foldr 实现了他自己的 ++ 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]和运行他们运行haskell，他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]

那么教授在他的推理中到底是什么意思？

顺便说一句，我正在尝试自己诱导如下：

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

等式右边的

foldr(:) ys xs表示对于这个Foldableys中的每个y，应用:到y和 xs，这实际上是左侧所做的。我的推理对吗？

Answer 1

IMO，您的推理是正确的 - 视频中有错误：

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

真的应该

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

如果您查看 video, you'll notice that Tom Lokhorst pointed out this error (among several others) 的评论。

`foldr` 的这个 `++` 实现是如何工作的？

How does this `++` implementation by `foldr` work?

haskell

functional-programming

fold