查找Java字节数组的缓存行的开始位置

对于高性能阻塞的Bloom过滤器,我希望将数据与缓存线对齐。(我知道在C中这样做更容易,但是我想使用Java。)

我确实有一个解决方案,但我不确定它是正确的,还是有更好的方法。我的解决方案尝试使用以下算法查找缓存线的起始位置:

• 对于每个可能的偏移量o(0..63;我假设缓存线长度为64)
• 启动从数据[O]读取并将其写入数据[O+8]的线程
• 在主线程中,将“1”写入数据[O],然后等待直到数据[O+8]结束(因此,等待另一个线程)
• 重复那个

然后,测量这有多快,基本上一个100万循环(在每个线程中)的增量是多少。我的逻辑是,如果数据在不同的缓存线中,速度会变慢。

这里是我的代码:

public static void main(String... args) {
    for(int i=0; i<20; i++) {
        int size = (int) (1000 + Math.random() * 1000);
        byte[] data = new byte[size];
        int cacheLineOffset = getCacheLineOffset(data);
        System.out.println("offset: " + cacheLineOffset);
    }
}

private static int getCacheLineOffset(byte[] data) {
    for (int i = 0; i < 10; i++) {
        int x = tryGetCacheLineOffset(data, i + 3);
        if (x != -1) {
            return x;
        }
    }
    System.out.println("Cache line start not found");
    return 0;
}

private static int tryGetCacheLineOffset(byte[] data, int testCount) {
    // assume synchronization between two threads is faster(?)
    // if each thread works on the same cache line
    int[] counters = new int[64];
    int testOffset = 8;
    for (int test = 0; test < testCount; test++) {
        for (int offset = 0; offset < 64; offset++) {
            final int o = offset;
            final Semaphore sema = new Semaphore(0);
            Thread t = new Thread() {
                public void run() {
                    try {
                        sema.acquire();
                    } catch (InterruptedException e) {
                        throw new RuntimeException(e);
                    }
                    for (int i = 0; i < 1000000; i++) {
                        data[o + testOffset] = data[o];
                    }
                }
            };
            t.start();
            sema.release();
            data[o] = 1;
            int counter = 0;
            byte waitfor = 1;
            for (int i = 0; i < 1000000; i++) {
                byte x = data[o + testOffset];
                if (x == waitfor) {
                    data[o]++;
                    counter++;
                    waitfor++;
                }
            }
            try {
                t.join();
            } catch (InterruptedException e) {
                throw new RuntimeException(e);
            }
            counters[offset] += counter;
        }
    }
    Arrays.fill(data, 0, testOffset + 64, (byte) 0);
    int low = Integer.MAX_VALUE, high = Integer.MIN_VALUE;
    for (int i = 0; i < 64; i++) {
        // average of 3
        int avg3 = (counters[(i - 1 + 64) % 64] + counters[i] + counters[(i + 1) % 64]) / 3;
        low = Math.min(low, avg3);
        high = Math.max(high, avg3);
    }
    if (low * 1.1 > high) {
        // no significant difference between low and high
        return -1;
    }
    int lowCount = 0;
    boolean[] isLow = new boolean[64];
    for (int i = 0; i < 64; i++) {
        if (counters[i] < (low + high) / 2) {
            isLow[i] = true;
            lowCount++;
        }
    }
    if (lowCount != 8) {
        // unclear
        return -1;
    }
    for (int i = 0; i < 64; i++) {
        if (isLow[(i - 1 + 64) % 64] && !isLow[i]) {
            return i;
        }
    }
    return -1;
}

它打印(示例):

offset: 16
offset: 24
offset: 0
offset: 40
offset: 40
offset: 8
offset: 24
offset: 40
...

因此,Java中的数组似乎与8字节对齐。