在java的线程概念中同步

Use*_*902 0 java multithreading synchronized

当我尝试执行这段代码时,我得到的输出就像12221121212121221212.当我声明方法时synchronized,它不应该允许其他线程接管控件

package com.test;

public class TestProject extends Thread {

    public void run() {
        synchronized (this) {
            for (int i = 0; i < 10; i++) {
                System.out.print(getName());
            }
        }
    }

    public static void main(String[] args) throws InterruptedException {
        TestProject myProject = new TestProject();
        TestProject myProject1 = new TestProject();
        myProject.setName("1");
        myProject1.setName("2");
        myProject.start();
        myProacject1.start();
    }
}
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那么,如何更改代码以获得顺序格式的输出1111111111222222222222222222221111111111

NPE*_*NPE 6

您正在同步this,即在TestProject对象上.由于您有两个单独的对象,因此您有两个单独的锁.因此,无论synchronized其他线程正在做什么,两个线程都可以进入该部分.

要跨实例同步,TestProject需要静态锁定:

public class TestProject extends Thread {

    private static final Object lock = new Object();

    public void run() {
        synchronized (lock) {
            ...
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  • +1或者`synchronized(TestProject.class)`. (2认同)