Java NIO Pipe 与 BlockingQueue

Question

我刚刚发现它只有一个 NIO 工具，即 Java NIO Pipe，它是为在线程之间传递数据而设计的。与通过队列（例如 ArrayBlockingQueue）传递的更传统的消息相比，使用此机制是否有任何优势？

Answer 1

通常，将数据传递给另一个线程进行处理的最简单方法是使用 ExecutorService。这封装了一个队列和一个线程池（可以有一个线程）。

当您有一个支持 NIO 通道的库时，您可以使用 Pipe。如果您想在线程之间传递数据的 ByteBuffer，它也很有用。

否则，使用 ArrayBlockingQueue 通常会更简单/更快。

如果您想要一种更快的方式在线程之间交换数据，我建议您查看 Exchanger 但它不像 ArrayBlockingQueue 那样通用。

Exchanger 和无 GC Java

Answer 2

我相信 NIO Pipe 的设计是为了让您可以以线程安全的方式将数据发送到选择器循环内的通道，换句话说，任何线程都可以写入管道，并且数据将在管道的另一个极端进行处理，在选择器循环内。当您写入管道时，您可以使另一端的通道可读。

Answer 3

因此，在管道遇到很多麻烦之后（查看此处）我决定支持非阻塞并发队列而不是 NIO 管道。所以我对Java的ConcurrentLinkedQueue做了一些基准测试。见下文：

public static void main(String[] args) throws Exception {

    ConcurrentLinkedQueue<String> queue = new ConcurrentLinkedQueue<String>();

    // first test nothing:

    for (int j = 0; j < 20; j++) {

        Benchmarker bench = new Benchmarker();

        String s = "asd";

        for (int i = 0; i < 1000000; i++) {
            bench.mark();
            // s = queue.poll();
            bench.measure();
        }

        System.out.println(bench.results());

        Thread.sleep(100);
    }

    System.out.println();

    // first test empty queue:

    for (int j = 0; j < 20; j++) {

        Benchmarker bench = new Benchmarker();

        String s = "asd";

        for (int i = 0; i < 1000000; i++) {
            bench.mark();
            s = queue.poll();
            bench.measure();
        }

        System.out.println(bench.results());

        Thread.sleep(100);
    }

    System.out.println();

    // now test polling one element on a queue with size one

    for (int j = 0; j < 20; j++) {

        Benchmarker bench = new Benchmarker();

        String s = "asd";
        String x = "pela";

        for (int i = 0; i < 1000000; i++) {
            queue.offer(x);
            bench.mark();
            s = queue.poll();
            bench.measure();
            if (s != x) throw new Exception("bad!");
        }

        System.out.println(bench.results());

        Thread.sleep(100);
    }

    System.out.println();

    // now test polling one element on a queue with size two

    for (int j = 0; j < 20; j++) {

        Benchmarker bench = new Benchmarker();

        String s = "asd";
        String x = "pela";

        for (int i = 0; i < 1000000; i++) {
            queue.offer(x);
            queue.offer(x);
            bench.mark();
            s = queue.poll();
            bench.measure();
            if (s != x) throw new Exception("bad!");
            queue.poll();
        }

        System.out.println(bench.results());

        Thread.sleep(100);
    }
}

结果：

totalLogs=1000000, minTime=0, maxTime=85000, avgTime=58.61 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=5281000, avgTime=63.35 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=725000, avgTime=59.71 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=25000, avgTime=58.13 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=378000, avgTime=58.45 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=15000, avgTime=57.71 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=170000, avgTime=58.11 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=1495000, avgTime=59.87 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=232000, avgTime=63.0 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=184000, avgTime=57.89 (times in nanos)

totalLogs=1000000, minTime=0, maxTime=2600000, avgTime=65.22 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=850000, avgTime=60.5 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=150000, avgTime=63.83 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=43000, avgTime=59.75 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=276000, avgTime=60.02 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=457000, avgTime=61.69 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=204000, avgTime=60.44 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=154000, avgTime=63.67 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=355000, avgTime=60.75 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=338000, avgTime=60.44 (times in nanos)

totalLogs=1000000, minTime=0, maxTime=345000, avgTime=110.93 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=396000, avgTime=100.32 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=298000, avgTime=98.93 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=1891000, avgTime=101.9 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=254000, avgTime=103.06 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=1894000, avgTime=100.97 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=230000, avgTime=99.21 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=348000, avgTime=99.63 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=922000, avgTime=99.53 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=168000, avgTime=99.12 (times in nanos)

totalLogs=1000000, minTime=0, maxTime=686000, avgTime=107.41 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=320000, avgTime=95.58 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=248000, avgTime=94.94 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=217000, avgTime=95.01 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=159000, avgTime=93.62 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=155000, avgTime=95.28 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=106000, avgTime=98.57 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=370000, avgTime=95.01 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=1836000, avgTime=96.21 (times in nanos)
totalLogs=1000000, minTime=0, maxTime=212000, avgTime=98.62 (times in nanos)

结论：

maxTime 可能很可怕，但我认为可以安全地得出结论，我们轮询并发队列的时间处于 50 纳秒范围内。

Answer 4

我认为管道将具有更好的延迟，因为它很可能通过幕后的协程来实现。因此，当数据可用时，生产者立即让步给消费者，而不是当线程调度程序决定时。

管道通常代表消费者-生产者问题，并且很可能以这种方式实现，以便两个线程协作并且不会被外部抢占。