JMH微基准测试递归快速排序

Question

你好，我正在尝试对各种排序算法进行微基准测试，我在jmh和基准快速排序方面遇到了一个奇怪的问题。也许我的执行有问题。如果有人能帮我看看问题在哪里，我会很感兴趣的。首先，我使用ubuntu 14.04, jdk 7和jmh 0.9.1。下面是我做基准测试的方法:

@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();
}
}

还有其他算法，但我把它们省略了，因为它们或多或少是可以的。由于某些原因，快速排序非常慢。时间变慢了!甚至更多-我需要分配更多的堆栈空间，它运行没有StackOverflowException。看起来出于某种原因，快速排序只是做了很多递归调用。有趣的是，当我简单地在我的主类中运行算法时，它运行得很好(具有相同的随机分布和100000个元素)。不需要增加堆栈和简单的纳米时间基准测试显示的时间非常接近其他算法。在基准测试中，当使用jmh进行测试时，JDK的排序非常快，并且与其他使用纳米时间基准测试的算法更加一致。是我做错了什么还是漏掉了什么?下面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，由于我的枢轴选择，如果我在已经排序的数据上运行，我的算法将非常慢(O(n^2))。但我仍然在随机数据上运行它，甚至当我在主方法中尝试在排序数据上运行它时，它比在随机数据上使用jmh的版本快得多。我很确定我遗漏了什么。您可以在这里找到其他算法的完整项目:https://github.com/ignl/SortingAlgos/

Answer 1

好吧，既然这里确实应该有一个答案(而不是必须浏览问题下面的评论)，我把它放在这里，因为我被烧伤了。

JMH中的迭代是一批基准测试方法调用(取决于将迭代设置为多长时间)。因此，使用@Setup(Level.Iteration)将只在调用序列的开始处进行设置。由于在第一次调用之后对数组进行了排序，因此在随后的调用中会在最糟糕的情况下(已排序的数组)调用快速排序。这就是为什么要花这么长时间，否则就会炸了。所以一个解决方案是使用@Setup(Level.Invocation)。但是，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Aleksey Shipilev建议，将数组复制成本吸收到每个基准方法中。因为您是在比较相对性能，所以这不应该影响您的结果。