Question

I just discovered that just has an NIO facility, Java NIO Pipe that s designed for passing data between threads. Is there any advantage of using this mechanism over the more conventional message passing over a queue, such as an ArrayBlockingQueue?

Answer 1

通常,通过数据供另一个线索处理的最简单办法是使用一种行政服务。这既包括一个电梯,又包括一个透水池(能够有一个透镜)。

如果有图书馆支持国家情报和安全局的渠道,你就可以使用管道。如果你想在校对之间传送数据,也是有益的。

否则,使用ArrayBlocking Queue的通常是简单/秘密的。

如果你想更快地在校对之间交换数据,我建议你研究Exchangeer 。然而,这并非一般目的,而是作为阿雷拉·邦马克。

交易所和GC-less Java

Answer 2

我认为,设计了一个NIO管道,以便你能够以可怕的安全方式将数据传送到选择方内的一个渠道,换言之,任何透镜都可以写到管道上,数据将在选择方圈内的其他极端处理。当你写给管道时,你就会把这个渠道放在另一个旁读。

Answer 3

因此,在管道发生许多麻烦之后()。我决定赞成不要在NIO管道上同时排队。因此,我对Java s Conjointed Queue做了一些基准。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:

最长的投票时间可以是平常的,但我认为,我们能够在50个南半球地区同时投票。

Answer 4

我相信管道会更加贴切,因为管道很可能在幕后安装cor。因此,当有数据时,生产者立即向消费者投产,而不是当阅读机决定时。

总的说来,管道是一个消费者-生产者的问题,很有可能以这种方式加以实施,以便二者合作,而不是外部预先防范。

友情链接