为什么两个Java线程(在某些情况下)的速度是一个的两倍？

Question

文件:Example1.java

public class Example1 implements Runnable {

    public void run() {
        for(int i = 0; i < 100000000; i++) {
            int x = 5;
            x = x * 4;
            x = x % 3;
            x = x + 9000;
            x = x * 923;
        }
    }

    public static void task() {
        for(int i = 0; i < 100000000; i++) {
            int x = 5;
            x = x * 4;
            x = x % 3;
            x = x + 9000;
            x = x * 923;
        }
        for(int i = 0; i < 100000000; i++) {
            int x = 9;
            x = x * 2;
            x = x % 4;
            x = x + 3241;
            x = x * 472;
        }
    }

    public static void main(String[] args) {

        long startTime = System.currentTimeMillis();
            Example1.task();
            Example1.task();
            Example1.task();
            Example1.task();
            Example1.task();
        long stopTime = System.currentTimeMillis();
        long runTime = stopTime - startTime;
        System.out.println("Run time for one thread: " + runTime);


        startTime = System.Example1();
            (new Thread(new Example1())).start();
            (new Thread(new Example2())).start();
            (new Thread(new Example1())).start();
            (new Thread(new Example2())).start();
            (new Thread(new Example1())).start();
            (new Thread(new Example2())).start();
            (new Thread(new Example1())).start();
            (new Thread(new Example2())).start();
            (new Thread(new Example1())).start();
            (new Thread(new Example2())).start();
        stopTime = System.currentTimeMillis();
        runTime = stopTime - startTime;
        System.out.println("Run time for two threads: " + runTime);


    }

}

文件:Example2.java

public class Example2 implements Runnable {

    public void run() {
        for(int i = 0; i < 100000000; i++) {
            int x = 9;
            x = x * 2;
            x = x % 4;
            x = x + 3241;
            x = x * 472;
        }        
    }
}

当我运行它时,它输出:

一个线程的运行时间:1219

两个线程的运行时间:281

或者非常接近的东西.

为什么会有这样的差异？为什么将它分成两个线程比直接运行它快两倍？

Answer 1

你实际上并没有等待线程完成.

一旦启动一个线程,就必须在其上调用.join()以等待它完成.这里发生的是你的所有线程都在启动,一旦最后一个线程启动,你计时它然后计算停止时间.这意味着您的线程仍在后台运行.

编辑:第一个需要这么长时间的原因是因为你正在进行一系列同步调用,而创建一个线程并启动它会产生一个异步任务.

编辑2:这是第一次测试中发生的情况的餐巾序列图:http://www.websequencediagrams.com/cgi-bin/cdraw？lz = TWFpbi0-RXhhbXBsZTE6IFRhc2sgc3RhcnRlZAphY3RpdmF0ZSAAGAgKACEILS0-TWFpbjogZG9uZQpkZQAYEgABWAABWAABgTFlMQo&s=napkin

这里是你的第二个测试会发生什么餐巾纸序列图: http://www.websequencediagrams.com/cgi-bin/cdraw?lz=TWFpbi0tPkFub255bW91cyBUaHJlYWQ6IFN0YXJ0IEV4YW1wbGUxLnRhc2soKQoACSYyAAEuAAFdAAGBOwCCPjoAgyIGPk1haW46ICJIb3cgbG9uZyBkaWQgdGhhdCB0YWtlPyIKAINmEC0AKwhUYXNrcyBiZWdpbiB0byBmaW5pc2guLi4gKHNvbWUgbWF5IGhhdmUgZW5kZWQgZWFybGllcikK&s=napkin

编辑3:我刚刚意识到第二个序列图将所有箭头指向/ same/thread.它们实际上是不同的线程,每次调用.