在java中给出了默认的新线程名称？

当我运行这个程序

public class Fabric extends Thread {
    public static void main(String[] args) {
        Thread t1 = new Thread(new Fabric());
        Thread t2 = new Thread(new Fabric());
        Thread t3 = new Thread(new Fabric());
        t1.start();
        t2.start();
        t3.start();
    }
    public void run() {
        for(int i = 0; i < 2; i++)
        System.out.print(Thread.currentThread().getName() + " ");
    }
}

我得到输出

Thread-1 Thread-5 Thread-5 Thread-3 Thread-1 Thread-3

是否有任何具体的原因，为什么线程被赋予奇数的名称 - 1,3,5 ......还是它是不可预测的？

---

new Thread(new Fabric());

由于Fabric是一个线程，所以你在这里创建了2个线程:)

JDK8代码：

/* For autonumbering anonymous threads. */
private static int threadInitNumber;
private static synchronized int nextThreadNum() {
    return threadInitNumber++;
}

---

除非在创建线程时指定名称，否则线程名称中的默认数值是递增值。 Fabric扩展了Thread ，并且您正在传递Fabric实例以创建另一个线程 - 因此内部线程计数器会在进程中创建2个线程时递增两次。

---

如果您更改程序，如下所示，您将按顺序获得线程编号。

    public class Fabric extends Thread {
    public static void main(String[] args) {
        Thread t1 = new Fabric();
        Thread t2 = new Fabric();
        Thread t3 = new Fabric();
        t1.start();
        t2.start();
        t3.start();
    }
    public void run() {
        for(int i = 0; i < 2; i++)
            System.out.print(Thread.currentThread().getName() + " ");
    }
}

和输出是

Thread-0 Thread-2 Thread-2 Thread-1 Thread-0 Thread-1

链接地址: http://www.djcxy.com/p/92115.html

上一篇: How is default new thread name given in java?

下一篇: Use of thread in Java class without implementing Runnable nor extending Thread