Haskell foldr具有功能
[英]Haskell foldr with function
问题描述
I would like to understand how the below code of Haskell foldr is evaluated. 
我想了解如何评估下面的Haskell foldr代码。
k x y = x
foldr k 1 [0..5]
The result is 0, but I can't understand why it is zero? 结果是0，但我不明白为什么它是零？ I would like to think that x is 1 any the elements in the list are y. 我想x是1，列表中的任何元素都是y。 Can anyone explain it to me, please? 请问有人向我解释一下吗？ I searched it online but couldn't found anything useful. 我在网上查了一下但找不到任何有用的东西。
1 个解决方案
解决方案1
10 已采纳 2015-03-12 15:08:30
The Haskell Wiki has some useful info about how to interpret foldr , including this image: Haskell Wiki有一些关于如何解释foldr有用信息，包括这个图像：
You can see how your expression expands to: 您可以看到表达式如何扩展为：
0 `k` (1 `k` ... (5 `k` 1)))))
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