具有等待和通知功能的Java多线程

Question

I have this piece of code

public MultiThreadedSum(ArrayBuffer ArrayBufferInst)
{
    this.ArrayBufferInst = ArrayBufferInst;
    Sum = 0;
    Flag = false;
    StopFlag = false;
}

public synchronized void Sum2Elements()
{

    while(Flag)
    {
        try {wait();}
        catch (InterruptedException e){}
    }

    Flag = true;

    if (StopFlag)
    {
        notifyAll();
        return;
    }

    System.out.println("Removing and adding 2 elements.");

    Sum = ArrayBufferInst.Sum2Elements();

    notifyAll();
}

public synchronized void InsertElement()
{

    while(!Flag)
    {
        try {wait();}
        catch (InterruptedException e){}
    }

    Flag = false;

    if (StopFlag)
    {
        notifyAll();
        return;
    }

    System.out.println("Inserting the sum.");

    ArrayBufferInst.InsertElement(Sum);

    if (ArrayBufferInst.RetunrSize() == 1)
    {
        StopFlag = true;
    }

    System.out.println(ArrayBufferInst);

    notifyAll();
}

As you can see, I set the Flag to be false first so one of the threads can enter the Sum2Elements method and change it to true and by that, making everyone wait.

I know that in synchronized code, only one thread can do its thing, well here I have two synchronized methods, does it mean that 2 threads are trying to conduct this methods after each notifyall?

And if so, is it not possible for one thread to enter Sum2Elements, change the flag to true before the other thread enters InsertElement, and by that skipping the while loop?

Thanks

Answer 1

Only one thread can hold the lock of the object. And then it's only that thread that can enter the synchronized methods on that object.

The thread can however release the lock without returning from the method, by calling Object.wait().

So your code looks good!

does it mean that 2 threads are trying to conduct this methods after each notifyall? 
Ans : It is very much possible for two threads to be in two of your synchronized methods since you are calling wait().

is it not possible for one thread to enter Sum2Elements, change the flag to true before the other thread enters InsertElement, and by that skipping the while loop?
Ans : Yes this is possible again for the same reason specified above.

Answer 2

Only one thread can execute one of two method at a time because both are synchronized though order is undefined

就像我说的那样，一个方法只能一次由一个线程执行，除非执行线程通过调用wait方法释放lock ，而另一个线程获得该lock并执行其他synchronized方法则使这两个语句成为可能 。

Answer 3

Locks are obtained on objects of a class & not on any particular synchronized method. Both the methods are instance methods. So if one of the threads have entered any synchronized method for an object, A say, then any other thread cant enter any synchronized method for that object until the running thread doesnt call notifyAll() method. At that stage all the waiting threads compete to become active but it depends on the thread scheduler to choose a thread which is to become active.

If you want that two different threads should access these synchronized methods simultaneously then the 2 threads should operate on 2 different objects of the class.