哈斯克尔函数折叠器 [英] Haskell foldr with function

问题描述

  kxy = x 
 $ 
我想了解下面的代码是如何评估Haskell foldr的。 b $ b foldr k 1 [0..5] 
  

结果为0，但我可以'不明白为什么它是零？我想认为x是1，列表中的任何元素都是y。任何人都可以向我解释一下吗？我在网上搜索了它，但找不到任何有用的东西。

Haskell Wiki 有关于如何解释 foldr 的一些有用信息，包括这张图片：

$ b $你可以看到你的表达式是如何展开的：

$ $ $ $ $ c $ 0 k`（1`k` ...（ 5`k` 1）））））

I would like to understand how the below code of Haskell foldr is evaluated.

k x y = x

foldr k 1 [0..5]

The result is 0, but I can't understand why it is zero? I would like to think that x is 1 any the elements in the list are y. Can anyone explain it to me, please? I searched it online but couldn't found anything useful.

解决方案

The Haskell Wiki has some useful info about how to interpret foldr, including this image:

You can see how your expression expands to:

0 `k` (1 `k` ... (5 `k` 1)))))

这篇关于哈斯克尔函数折叠器的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！