multiple synchronized block in class..

[vipin dwivedi]

Hi all,
I have a small doubt any one can clearit ...
if a class have multiple Synch. methods or blocks ,then only one lock dfined for that class or multiple locks are defined for each synch. block.
//dummy code
class a{
synchronized m1(){
m2();
}
synchronized m2(){
}
}
Now for the same class if thread t1 calles m1() ,and t2 calles m2()..Then is it possible to call m2() by t2 untill m1() exits for t1.
practically its working ..... my question is how it is working ,by Scheduling ?

[Ernest Friedman-Hill]

I don't understand your last sentence or two about what's happening or what you expect to happen, but I can tell you that there is just one lock per object, and that while one thread is running in m1() or m2(), no other thread will be able to enter either of these methods.

[Vinod Chandana]

Originally posted by vipin dwivedi:
Hi all,
I have a small doubt any one can clearit ...
if a class have multiple Synch. methods or blocks ,then only one lock dfined for that class or multiple locks are defined for each synch. block.
//dummy code
class a{
synchronized m1(){
m2();
}
synchronized m2(){
}
}
Now for the same class if thread t1 calles m1() ,and t2 calles m2()..Then is it possible to call m2() by t2 untill m1() exits for t1.
practically its working ..... my question is how it is working ,by Scheduling ?

Hi Vipin,
Ur question is a bit vague. Let me see if I can answer ur question. If t1 and t2 try to enter m1 and m2 at the same time, then they will be able to enter both the methods. But thread t1 has to wait until t2 exits m2 since m2 is already being used by t2 and t2 holds the lock.
Hope this helps,
Vinod.

[Ernest Friedman-Hill]

If t1 and t2 try to enter m1 and m2 at the same time, then they will be able to enter both the methods. But thread t1 has to wait until t2 exits m2 since m2 is already being used by t2 and t2 holds the lock.

Sorry, but this is wrong. There is only one lock which covers both methods. If thread t2 is already running in m2, then thread t1 will not be able to begin running m1 until m2 returns in t2.

[Vinod Chandana]

I am a little surprised. But have a look at the code and the output. This is contradicts ur explaination.
Thanks,
Vinod.

public class Test {
    public synchronized void m1(){
        System.out.println("In m1 at " + System.currentTimeMillis());
        try{
            
            wait(20);
        }catch(Exception e){
            e.printStackTrace();
        }
        m2();
        System.out.println("Exiting m1 at " + System.currentTimeMillis());
    }
    
    public synchronized void m2(){
        System.out.println("In m2 at " + System.currentTimeMillis());
        try{
            wait(40);
        }catch(Exception e){
            e.printStackTrace();
        }
        System.out.println("Exiting m2 at " + System.currentTimeMillis());
    }
}
public class ThreadTest {
    public static void main(String[] args) {
       MyThread t1 = new MyThread(1);
       MyThread t2 = new MyThread(2);
       t1.start();
       t2.start();
    }
    
}
public class MyThread extends Thread{
    int id;
    static Test myObj = new Test();
    public MyThread(int i) {
        this.id = i;
    }
    
    public void run(){
        if(id % 2 == 0) myObj.m2();
        else myObj.m1();
    }
    
}
Output:
=======
In m1 at 1066769163718
In m2 at 1066769163718
In m2 at 1066769163765
Exiting m2 at 1066769163765
Exiting m2 at 1066769163812
Exiting m1 at 1066769163812

Originally posted by Ernest Friedman-Hill:

Sorry, but this is wrong. There is only one lock which covers both methods. If thread t2 is already running in m2, then thread t1 will not be able to begin running m1 until m2 returns in t2.

[Ernest Friedman-Hill]

But have a look at the code and the output. This is contradicts ur explaination.

No, not at all. In the original example, there were no wait() calls. wait() changes everything! The wait() method can only be called from inside a synchronized method or block, and it explicitly surrenders the lock that the synchronized block acquired. At this point, another Thread that may have been waiting to grab the lock is free to proceed. wait() can't return until that second thread has released the lock. This is all described in the Javadocs for java.lang.Object.
Maybe you meant to use Thread.sleep() instead of wait()? In any case,if the wait() calls were changed to Thread.sleep() calls, you'd find that the output changes dramatically; during a sleep() call, a Thread retains all the locks that it held before the call.

[Vinod Chandana]

Got it.. U r the man.
- Vinod.