java - 对象的成员不需要 volatile 而只需要原始成员吗？

Question

我的代码是

package threadrelated;
import threadrelated.lockrelated.MyNonBlockingQueue;

public class VolatileTester extends Thread {

 MyNonBlockingQueue mbq ;

 public static void main(String[] args) throws InterruptedException {

    VolatileTester vt = new VolatileTester();
    vt.mbq = new MyNonBlockingQueue(10);
    System.out.println(Thread.currentThread().getName()+" "+vt.mbq);
    Thread t1 = new Thread(vt,"First");
    Thread t2 = new Thread(vt,"Secondz");
    t1.start();
    t2.start();
    t1.join();
    t2.join();
    System.out.println(Thread.currentThread().getName()+" "+vt.mbq);

}
@Override
public void run() {
    System.out.println(Thread.currentThread().getName()+" before    "+mbq);
    mbq = new MyNonBlockingQueue(20);
    try {
        Thread.sleep(TimeUnit.SECONDS.toMillis(10));
    } catch (InterruptedException e) {
        e.printStackTrace();
    }
    System.out.println(Thread.currentThread().getName()+" after   "+mbq);
}

}

输出是

main threadrelated.lockrelated.MyNonBlockingQueue@72fcb1f4
Secondz before    threadrelated.lockrelated.MyNonBlockingQueue@72fcb1f4
First before    threadrelated.lockrelated.MyNonBlockingQueue@72fcb1f4
Secondz after   threadrelated.lockrelated.MyNonBlockingQueue@7100650c
First after   threadrelated.lockrelated.MyNonBlockingQueue@7100650c
main threadrelated.lockrelated.MyNonBlockingQueue@7100650c

说明当First线程将成员变量赋值给new对象时，同样对其他线程可见。即使“mbq”未声明为易变的。

我使用断点来尝试不同的操作顺序。但我的观察是一个线程可以立即看到另一个线程的影响。

类成员 object 不需要 volatile 吗？它们总是与主存同步吗？仅原始成员变量(int、long、boolean 等？)需要 volatile

Answer 1

它对引用和基元一样必要。您的输出没有显示可见性问题这一事实并不证明不存在。一般来说，很难证明不存在并发错误。但这里有一个简单的反证，显示了 volatile 的必要性:

public class Test {
    static volatile Object ref;

    public static void main(String[] args) {
        // spin until ref is updated
        new Thread(() -> {
            while (ref == null);
            System.out.println("done");
        }).start();

        // wait a second, then update ref
        new Thread(() -> {
            try { Thread.sleep(1000); } catch (Exception e) {}
            ref = new Object();
        }).start();
    }
}

该程序运行一秒钟，然后打印“完成”。删除 volatile 并且它不会终止，因为第一个线程永远不会看到更新的 ref 值。 (免责声明:与任何并发测试一样，结果可能会有所不同。)