Groovy：Closure如何在sort（）中工作 [英] Groovy: How Closure works in sort()

问题描述

我找到了一个代码段：

I found a code segment like:

def l1 = ["hello","hi","hey"]    
l1.sort{new Random()}

我无法弄清楚Random类对象是usd排序/洗牌列表项目？如何使用Random类对象返回Comparable / Comparator对象来执行操作？

I could not figure out how a Random class object is being usd to sort/shuffle the list items? How Random class object is returning Comparable/Comparator object to perform operation?

在groovy文档中，我找到了以下语句：
使用给定的Closure对此Iterable进行排序以确定正确的顺序。
但在这里如何使用Random？

In the groovy doc, I found the statement: Sorts this Iterable using the given Closure to determine the correct ordering. But how Random is used here?

推荐答案

我怀疑这里的意图是随机值被用作每个条目的 Comparable （即使用 Iterable.sort（Closure closure） ）。正在进行比较的是这些随机值，因此决定了排序顺序。因此，这导致了一个随机的排序顺序。

I suspect that the intent here is that a random value is used as the Comparable for each entry (i.e. using the single-argument closure variant of Iterable.sort(Closure closure)). It is these random values that are being compared, and thus dictating the sort order. Thus this leads to a randomised sort order.

然而，我不相信这会导致明确的行为。该文档没有定义：

However, I'm not convinced this leads to well-defined behaviour. The documentation does not define whether:

1. 对于迭代器的每个元素，闭包都被精确计算一次。


2. sort 应如何响应不一致的可比较值。

1. The closure is evaluated precisely once for each element of the iterable.
2. How sort should respond to inconsistent Comparable values.

如果＃1不是真的，那么会产生不一致的值 - 目前还不清楚这是否会导致 sort 实现变得古怪事情。

If #1 is not true, then inconsistent values would get generated - it's unclear whether that would cause the sort implementation to do wacky things.

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