haskell - 在这个 haskell 示例中 loeb 如何终止？

Question

来自 this Youtube video :

> let loeb fs = xs where xs = fmap ($ xs) fs
> loeb [length, (!! 0)]
[2,2]

xs这里是递归定义的，而 loeb 如何终止超出了我的范围。

Answer 1

尝试一下:

loeb [length, (!! 0)]
= xs where xs = fmap ($ xs) [length, (!! 0)]
= xs where xs = [length xs, xs !! 0]

所以当然 length xs只是 2，所以 xs 的第一个元素( length xs ) 也是 2。

记住: length xs不需要评估列表的项目:

length [undefined, undefined] = 2