懒惰的评估如何运作？(How does lazy evaluation work?)

 我是函数式编程的新手，我想要了解懒惰的评估是如何工作的。 我有以下功能： 
 
nats = 0:(map (+1) nats)
test = foldr (\x y-> if x > 2 then 0 else x+y) 10 nats
 
 据我所知，第一个得到所有自然数的列表，直到它耗尽内存（但从来没有真正发生过，我明白），而第二个应该是做什么的？ 测试里面有一个lambda演算函数，它得到一个x和y，如果x> 2且x + y else，则返回2，但它是否意味着它将返回0，因为x = 10？ 我有点困惑。 在那种情况下，nats真的需要吗？ 
 
 编辑：测试返回3，但现在我真的很困惑，因为我不明白如何。 
 
 如果我犯了一些菜鸟错误，谢谢，对不起，但我不明白这是怎么回事。 

I'm new to functional programming and I want to undersatand how lazy evaluation works. I have the following functions:
 
nats = 0:(map (+1) nats)
test = foldr (\x y-> if x > 2 then 0 else x+y) 10 nats
 
And as far as I can tell, the first one gets the list of all natural numbers until it runs out of memory (but that never really happens, that I understand), and the second one is supposed to do exactly what? Test has inside a lambda calculus function that gets an x and y and returns 2 if x > 2 and x+y else, but is that supposed to mean that it will return 0 because x = 10? I'm a little confused. In that case, is nats really called for?
 
EDIT: Test returns 3, but now I'm really confused because I don't understand how.
 
Thanks and sorry if I'm making some rookie mistake, but I'm failing to understand how this works.

原文：https://stackoverflow.com/questions/43783128