Synchronization and Locks

[Steven Wong]

Hi,
I'm still a little bit confused about the concept of synchronization and locks.
For the below code, is there only one object holding a lock to the run method thus when
thread 1 runs, thread 2 cannot access the run method.
What's the difference if I replace the first implementation of the run method with the commented one?
I am not getting something here....Can someone pls help?

 
package com.efkon.test;
public class cn28 extends Thread
{
cn28(String name){
super(name);
}

public static void main(String[] args)
{
cn28 c1 = new cn28("Thread One");
cn28 c2 = new cn28("Thread Two");
c1.start();
c2.start();
}

public synchronized void run()//Line1
{
System.out.println(Thread.currentThread().getName() + " is running...");
}

/*public void run()//Line2
{
 synchronized(this)
{
System.out.println(Thread.currentThread().getName() + " is running...");
}
}*/
}
 

Thanks,
Clement

[Rob Ross]

In this example, there is no difference. Acquiring a lock via a synchronized block like this:

void foo(){
  synchronized (this){
     //code
   }
} //end of method

has the exact same effect as writing
synchronized void foo(){
//code
}
In both cases, you are specifying that only one thread can run the method/block at a time, because you are synchronizing on the object itself. In the case of the synchronized method, the "this" is implicit. It's explicit in the synchronized block. The benifit of using a block then, is that you can synchronize on some different object than "this", if your particular program needs to do that for a variety of reasons.
But like I said, in your program, they are functionally equivalent.

[Steven Wong]

So, in this case, are there two different objects holding a lock on to itself each time they
enter the run method ?
Thanks,
Clement

[Rob Ross]

No, there are two (or more) threads each trying to get the same lock, and since only one thread can get the lock at a time, only one thread will be able to run your run() method in your class.
[ May 15, 2002: Message edited by: Rob Ross ]

[Steven Wong]

Hi Rob,
Thanks for your fast response.
Pls have a look at the following code which displays a random value of i.
My question is, as you said only one thread can obtain a lock on the run method at a time,
why is it that the value of i is not printed in order, ie either Thread1 (or Thread2)
finish printing the value of i first, only then the next thread gets the turn.

 

package com.efkon.test;
public class cn28 extends Thread
{
cn28(String name){
super(name);
}

public static void main(String[] args)
{
cn28 c1 = new cn28("Thread One");
cn28 c2 = new cn28("Thread Two");
c1.start();
c2.start();
}

public synchronized void run()//Line1
{
for(int i=0;i<10;i++)
{
System.out.println(Thread.currentThread().getName() + " is running... at i=" +i);
}
}

/*public void run()//Line2
{
synchronized(this)
{
System.out.println(Thread.currentThread().getName() + " is running...");
}
}*/

}
 

Thanks,
Clement

[Steven Sun]

I think the key is here:
for synchronized method:
only the datas that accessed by the method will be locked,not means that the second thread cant invoke the same method.Situation can be thus:the two thread both can get chance to run,when thread1 get it,it prints something but not all,then thread2 get the chance to run,because the method didnt lock any datas,so thread2 can easily continue,so it also print.
may be :
Thread One is running... at i=0
Thread One is running... at i=1
Thread One is running... at i=2
Thread One is running... at i=3
Thread One is running... at i=4
Thread One is running... at i=5
Thread One is running... at i=6
Thread One is running... at i=7
Thread Two is running... at i=0
Thread Two is running... at i=1
Thread Two is running... at i=2
Thread Two is running... at i=3
Thread Two is running... at i=4
Thread Two is running... at i=5
Thread Two is running... at i=6
Thread Two is running... at i=7
Thread Two is running... at i=8
Thread Two is running... at i=9
Thread One is running... at i=8
Thread One is running... at i=9
i think the synchronized object also the same reason.
if i am wrong,correct me

[luco zhao]

I think there r two threads running?

[Steven Wong]

Hi,
1. I think it's because there are two instances of thread being created, c1 and c2
and the synchronization is being done on each of them ('this').
But I'm a bit unsure how many objects were created actually ?
for eg.
Thread t1 = new Thread();
Thread t2 = new Thread();
Are there two objects being created for the above? Hmm...going back to the basics...
2. Check out the below code, where the synchronization is done on the object o. It will print out
the value of i in order.

 
package com.efkon.test;
public class cn28 extends Thread
{
static cn28 c1 = new cn28("Thread One");
static cn28 c2 = new cn28("Thread Two");
static Object o = new Object();

cn28(String name)
{
super(name);
}
public static void main(String[] args)
{
c1.start();
c2.start();
}
public void run()
{
synchronized (o)
{
for (int i = 0; i < 10; i++)
{
    System.out.println("Value of i is " + i + " for thread " + getName());
}
}
}
}
 

Thanks,
Clement

[Steven Wong]

Hi,
Did not get a reply from anyone just yet.
Rob, Corey, bartenders, etc.?? Anyone to help out, pls?

Clement

[Steven Sun]

because Object o is locked by synchronized,so the second thread will wait for the object lock to be set free by the first one,so the output will be in order.
is that clear??
Object o is "public" for both threads,that is the key.

[Junilu Lacar]

Clement,
The key point to remember is that the locking is done on a specific object. The same code can still be executed concurrently by different threads if they are using two different instances.
In your first example where you synchronize the run method, the synchronization is pretty useless and just adds overhead.
// two objects created
MyThread t1 = new MyThread();
MyThread t2 = new MyThread();
// these two threads still execute the code
// in run() concurrently despite the synchronize
t1.start(); // executes run() with lock on t1
t2.start(); // executes run() with lock on t2
In your second example where you synchronize on a common object, the two threads have to wait to obtain the lock on o, which is why you are getting the results you expected.
See this JavaWorld article for a good discussion of threads.
HTH,
Junilu

[Steven Wong]

Got it! Thanks for the help, guys.
Clement

[Francisco A Guimaraes]

Hi, taking advantage of Clement�s doubt, if the class implemented Runnable instead of extending Thread, would the syncronization make a difference?
Thanks,
Francisco

[Corey McGlone]

Originally posted by Francisco A Guimaraes:
...if the class implemented Runnable instead of extending Thread, would the syncronization make a difference?

No. Locking and synchronization is done at the Object level (i.e. every object and class has a lock). Therefore, how synchronization is performed is not based upon what type of object you have (Thread vs. Runnable).
Corey

[Steven Sun]

Originally posted by Steven Sun:
I think the key is here:
for synchronized method:
only the datas that accessed by the method will be locked,not means that the second thread cant invoke the same method.Situation can be thus:the two thread both can get chance to run,when thread1 get it,it prints something but not all,then thread2 get the chance to run,because the method didnt lock any datas,so thread2 can easily continue,so it also print.
may be :
Thread One is running... at i=0
Thread One is running... at i=1
Thread One is running... at i=2
Thread One is running... at i=3
Thread One is running... at i=4
Thread One is running... at i=5
Thread One is running... at i=6
Thread One is running... at i=7
Thread Two is running... at i=0
Thread Two is running... at i=1
Thread Two is running... at i=2
Thread Two is running... at i=3
Thread Two is running... at i=4
Thread Two is running... at i=5
Thread Two is running... at i=6
Thread Two is running... at i=7
Thread Two is running... at i=8
Thread Two is running... at i=9
Thread One is running... at i=8
Thread One is running... at i=9
i think the synchronized object also the same reason.
if i am wrong,correct me

Sorry for my misunderstanding.
the key should be that u have start two threads ,that is, you have two cn28 objects,so synchronized is meaningless.Remember,synchronized is for one object,only u want to use multi-threads to change a single object or its members,synchronized is helpful.