寻求如何避免僵局的建议

Question

我的应用程序中有一个类似的情况：

public class Example {
private static A aClass = new A();
private static B bClass = new B();
public static void main(String[] args) {
    new Thread(new Runnable() {
        @Override
        public void run() {
            bClass.start();
        }
    }, ">>>>>>>>>>").start();

    new Thread(new Runnable() {
        @Override
        public void run() {
            aClass.start();
        }
    }, "__________").start();
    try {
        Thread.sleep(200);
    } catch (InterruptedException e) {
        e.printStackTrace();
    }
    detectDeadlock();
}
private static void detectDeadlock() {
    ThreadMXBean threadBean = ManagementFactory.getThreadMXBean();
    long[] threadIds = threadBean.findMonitorDeadlockedThreads();
    int deadlockedThreads = threadIds != null? threadIds.length : 0;
    System.out.println("Number of deadlocked threads: " + deadlockedThreads);
}
private static class A {
    private final Object aLock = new Object();
    public void start() {
        System.out.println(Thread.currentThread()+ " " + "A-start; waiting for aLock");
        synchronized (aLock) {
            System.out.println(Thread.currentThread()+ " " + "A-start; acquired aLock");
            bClass.synchronizedMethod();
        }
        System.out.println(Thread.currentThread()+ " " + "A-start; release aLock");
    }
    public void synchronizedMethod() {
        System.out.println(Thread.currentThread()+ " " + "A-synchronizedMethod; waiting for aLock");
        synchronized (aLock) {
            System.out.println(Thread.currentThread()+ " " + "A-synchronizedMethod; acquired aLock");
        }
        System.out.println(Thread.currentThread()+ " " + "A-synchronizedMethod; release aLock");
    }
}
private static class B {
    private final Object bLock = new Object();
    public void synchronizedMethod() {
        System.out.println(Thread.currentThread()+ " " + "B-synchronizedMethod; waiting for bLock");
        synchronized (bLock) {
            System.out.println(Thread.currentThread()+ " " + "B-synchronizedMethod; acquired bLock");
        }
        System.out.println(Thread.currentThread()+ " " + "B-synchronizedMethod; release bLock");
    }
    public void start() {
        System.out.println(Thread.currentThread() + " " + "B-Start; waiting for bLock");
        synchronized (bLock) {
            System.out.println(Thread.currentThread()+ " " + "B-Start; acquired bLock");
            try {
                Thread.sleep(100);
            } catch (InterruptedException e) {
                e.printStackTrace();
            }
            aClass.synchronizedMethod();
        }
        System.out.println(Thread.currentThread()+ " " + "B-Start; released bLock");
    }
}

}

这个片段的输出是这样的：

Thread[>>>>>>>>>>,5,main] B-Start; waiting for bLock
Thread[>>>>>>>>>>,5,main] B-Start; acquired bLock
Thread[__________,5,main] A-start; waiting for aLock
Thread[__________,5,main] A-start; acquired aLock
Thread[__________,5,main] B-synchronizedMethod; waiting for bLock
Thread[>>>>>>>>>>,5,main] A-synchronizedMethod; waiting for aLock
Number of deadlocked threads: 2

我正在为这种僵局寻求一些指导。

你能告诉我如何避免这种僵局吗？

Answer 1

最后，我选择使用一个全局锁，该锁将由a和B的两个实例共享。

代码看起来像这样：

 private static class Lock {
    private static Lock lock = new Lock();
    private Lock(){}
    public static Lock getLock() {return lock;}
}
private static class A {
    private final Object aLock = Lock.getLock();
    // ...
}
private static class B {
    private final Object bLock = Lock.getLock();
    // ...
}