多线程-同步代码块中的隐患及解决办法-Peak-Gao

多线程访问同步代码块出现阻塞并解决

/**
 * 多线程访问同步代码块出现阻塞并解决
 * 锁代码块：作用的对象时整个代码块、每个对象只有一个锁与之关联
 * 阻塞原因：synchronized会将当前对象锁住，只有执行完才会释放该对象锁，下一个线程才能得到锁，从而形成互斥性
 */
public class SynchronizedCodeQuick {

    public static void main(String[] args) {
        //多线程访问会阻塞 原因没释放锁，一个对象一个锁形成互斥性
//        MyThread myThread = new MyThread();
//        Thread thread1 = new Thread(myThread, "thread1");
//        Thread thread2 = new Thread(myThread, "thread2");
//        thread1.start();
//        thread2.start();
        //多线程访问不会阻塞 两把锁互不干扰、没有互斥性
        Thread thread1 = new Thread(new MyThread(),"thread1")
        Thread thread2 = new Thread(new MyThread(),"thread2")
        thread1.start()
        thread2.start()
    }

}


class MyThread implements Runnable {

    @Override
    public void run() {
        synchronized (this) {
            for (int i = 0 i < 5 i++) {
                System.out.println("线程名：" + Thread.currentThread().getName() + "：" + i)
                try {
                    Thread.sleep(100)
                } catch (InterruptedException e) {
                    e.printStackTrace()
                }
            }
        }
    }
}

注释的代码运行如下：

优化后的效果

多线程访问synchronized和非synchronized

/**
 * 多线程访问synchronized和非synchronized 并不阻塞,证实了synchronized作用范围就是被修饰的代码块
 */
public class IsBlockSynchronizedCodeQuick {
    public static void main(String[] args) {
        MyIsThread myThread = new MyIsThread()
        Thread thread1 = new Thread(myThread, "thread1")
        Thread thread2 = new Thread(myThread, "thread2")
        thread1.start()
        thread2.start()
    }
}

class MyIsThread implements Runnable {

    public void test1() {
        synchronized (this) {
            for (int i = 0 i < 5 i++) {
                System.out.println("线程名" + Thread.currentThread().getName() + ":" + i)
                try {
                    Thread.sleep(100)
                } catch (InterruptedException e) {
                    e.printStackTrace()
                }
            }
        }
    }

    public void test2() {
        for (int i = 0 i < 5 i++) {
            System.out.println("线程名" + Thread.currentThread().getName() + ":" + i)
            try {
                Thread.sleep(100)
            } catch (InterruptedException e) {
                e.printStackTrace()
            }
        }
    }


    @Override
    public void run() {
        String name = Thread.currentThread().getName()
        if("thread1".equals(name)){
            test1()
        }else {
            test2()
        }
    }
}

总结：