java JMH微基准递归快速排序

Question

您好，我正在尝试对各种排序算法进行微基准测试，我发现jmh和基准测试quicksort存在一个奇怪的问题。也许我的实现有问题。如果有人能帮我看看问题出在哪里，我会很感兴趣的。首先，我将ubuntu 14.04与jdk 7和jmh 0.9.1结合使用。 以下是我如何尝试进行基准测试：

@OutputTimeUnit(TimeUnit.MILLISECONDS)
@BenchmarkMode(Mode.AverageTime)
@Warmup(iterations = 3, time = 1)
@Measurement(iterations = 3, time = 1)
@State(Scope.Thread)
public class SortingBenchmark {

private int length = 100000;

private Distribution distribution = Distribution.RANDOM;

private int[] array;

int i = 1;

@Setup(Level.Iteration)
public void setUp() {
    array = distribution.create(length);
}

@Benchmark
public int timeQuickSort() {
    int[] sorted = Sorter.quickSort(array);
    return sorted[i];
}

@Benchmark
public int timeJDKSort() {
    Arrays.sort(array);
    return array[i];
}

public static void main(String[] args) throws RunnerException {
    Options opt = new OptionsBuilder().include(".*" + SortingBenchmark.class.getSimpleName() + ".*").forks(1)
            .build();

    new Runner(opt).run();
}
}

还有其他的算法，但我把它们遗漏了，因为它们或多或少是好的。由于某些原因，现在快速排序速度非常慢。时间的大小变慢了！更重要的是，我需要分配更多的堆栈空间，以便它在没有StackOverflowException的情况下运行。出于某种原因，quicksort似乎只执行大量递归调用。有趣的是，当我在我的主类中简单地运行算法时，它运行得很好（具有相同的随机分布和100000个元素）。不需要增加堆栈，简单的nanotime基准测试显示的时间与其他算法非常接近。在基准测试中，当使用jmh进行测试时，JDK排序速度非常快，并且与其他使用naive nanotime基准测试的算法更加一致。我是做错了什么还是错过了什么？ 以下是我的快速排序算法：

public static int[] quickSort(int[] data) {
    Sorter.quickSort(data, 0, data.length - 1);
    return data;
}
private static void quickSort(int[] data, int sublistFirstIndex, int sublistLastIndex) {
    if (sublistFirstIndex < sublistLastIndex) {
        // move smaller elements before pivot and larger after
        int pivotIndex = partition(data, sublistFirstIndex, sublistLastIndex);
        // apply recursively to sub lists
        Sorter.quickSort(data, sublistFirstIndex, pivotIndex - 1);
        Sorter.quickSort(data, pivotIndex + 1, sublistLastIndex);
    }
}
private static int partition(int[] data, int sublistFirstIndex, int sublistLastIndex) {
    int pivotElement = data[sublistLastIndex];
    int pivotIndex = sublistFirstIndex - 1;
    for (int i = sublistFirstIndex; i < sublistLastIndex; i++) {
        if (data[i] <= pivotElement) {
            pivotIndex++;
            ArrayUtils.swap(data, pivotIndex, i);
        }
    }
    ArrayUtils.swap(data, pivotIndex + 1, sublistLastIndex);
    return pivotIndex + 1; // return index of pivot element
}

现在我明白了，由于我的轴选择，如果我在已经排序的数据上运行算法，那么我的算法将非常慢（O（n^2））。但我仍然在随机化的方法上运行它，甚至当我在我的主要方法中尝试在排序数据上运行它时，它比在随机化数据上使用jmh的版本快得多。我很确定我错过了什么。您可以在此处找到具有其他算法的完整项目：https://github.com/ignl/SortingAlgos/

Answer 1

好吧，既然这里真的应该有一个答案（而不是必须通过问题下面的评论），我把它放在这里，因为我被烧伤了

JMH中的迭代是一批基准方法调用（取决于迭代设置的时间）。因此，使用@Setup（Level.Iteration）只会在一系列调用开始时进行设置。由于数组是在第一次调用后进行排序的，所以在后续调用中，在最糟糕的情况下（排序数组）会调用quicksort。这就是为什么要花这么长时间，否则会把事情搞砸

因此，解决方案是使用@Setup（Level.Invocation）。然而，正如Javadoc中所述：

**
     * Invocation level: to be executed for each benchmark method execution.
     *
     * <p><b>WARNING: HERE BE DRAGONS! THIS IS A SHARP TOOL.
     * MAKE SURE YOU UNDERSTAND THE REASONING AND THE IMPLICATIONS
     * OF THE WARNINGS BELOW BEFORE EVEN CONSIDERING USING THIS LEVEL.</b></p>
     *
     * <p>This level is only usable for benchmarks taking more than a millisecond
     * per single {@link Benchmark} method invocation. It is a good idea to validate
     * the impact for your case on ad-hoc basis as well.</p>
     *
     * <p>WARNING #1: Since we have to subtract the setup/teardown costs from
     * the benchmark time, on this level, we have to timestamp *each* benchmark
     * invocation. If the benchmarked method is small, then we saturate the
     * system with timestamp requests, which introduce artificial latency,
     * throughput, and scalability bottlenecks.</p>
     *
     * <p>WARNING #2: Since we measure individual invocation timings with this
     * level, we probably set ourselves up for (coordinated) omission. That means
     * the hiccups in measurement can be hidden from timing measurement, and
     * can introduce surprising results. For example, when we use timings to
     * understand the benchmark throughput, the omitted timing measurement will
     * result in lower aggregate time, and fictionally *larger* throughput.</p>
     *
     * <p>WARNING #3: In order to maintain the same sharing behavior as other
     * Levels, we sometimes have to synchronize (arbitrage) the access to
     * {@link State} objects. Other levels do this outside the measurement,
     * but at this level, we have to synchronize on *critical path*, further
     * offsetting the measurement.</p>
     *
     * <p>WARNING #4: Current implementation allows the helper method execution
     * at this Level to overlap with the benchmark invocation itself in order
     * to simplify arbitrage. That matters in multi-threaded benchmarks, when
     * one worker thread executing {@link Benchmark} method may observe other
     * worker thread already calling {@link TearDown} for the same object.</p>
     */ 

因此，正如Aleksey Shipilev所建议的，将阵列拷贝成本吸收到每个基准方法中。因为你是在比较相对性能，所以这不应该影响你的结果