Java 7排序“优化” [英] Java 7 sorting "optimisation"

问题描述

在Java6中，quicksort和mergesort分别用于 Arrays＃sort ，分别用于原始和对象数组。在Java7中，这些都已经改变， DualPivotQuicksort 和Timsort。

In Java6 both quicksort and mergesort were used in Arrays#sort, for primitive and object arrays respectively. In Java7 these have both changed, to DualPivotQuicksort and Timsort.

在新快速排序的来源中，以下评论会出现在几个地方（例如第354行）：

In the source for the new quicksort, the following comment appears in a couple of places (eg line 354):

 /*
  * Here and below we use "a[i] = b; i++;" instead
  * of "a[i++] = b;" due to performance issue.
  */

这是一个性能问题？编译器不会将这些减少到同样的事情吗？

How is this a performance issue? Won't the compiler will reduce these to the same thing?

更广泛地说，自己调查这个的好策略是什么？我可以运行基准测试，但我更感兴趣的是分析编译代码中的任何差异。但是，我不知道使用什么工具等。

More broadly, what is a good strategy for investigating this myself? I can run benchmarks, but I'd be more interested in analysing any differences in the compiled code. However, I don't know what tools to use etc.

推荐答案

这只是一般问题的答案。

This is only an answer to the general question.

您可以查看字节码并尝试理解差异。即你可以用 a [i] = b;写一个简单的例子。 i ++; 和 a [i ++] = b; ，看看差异是什么。

You can look at the bytecode and try to understand the differences. I.e. you could write yourself a simple example using both a[i] = b; i++; and a[i++] = b; and see whats the difference.

显示字节码的最简单方法是 javap 程序（应该包含在JDK中）。使用 javac SomeFile.java 编译代码并在代码上运行javap： javap -c SomeFile （ - c开关告诉javap输出文件中每个方法的字节码。）

The simplest way to show the bytecode is the javap program (should be included in you JDK). Compile the code with javac SomeFile.java and run javap on the code: javap -c SomeFile (the -c switch tells javap to output the bytecode for each method in the file).

如果你正在使用eclipse，你也可以试试这一个。

If you are using eclipse you could also try this one.

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