Nested synchronized keyword

Question

I have this code in Java:

    public void doSomeThing() {
        synchronized (this) {
            doSomeThingElse();
        }
    }
    public void doSomeThingElse() {
        synchronized (this) {
            // do something else
        }
    }

Can this code block? I mean, Can this code wait for ever?

Kissaki · Accepted Answer · 2013-10-16T17:20:57.990

57

As the java documentation describes for Reentrant locking:

a thread can acquire a lock that it already owns

The second synchronized block is using the same lock and thus will always be useable as the lock has already been aquired in the outer method.

No, there will not be a deadlock.

edited Oct 16 '13 at 17:20

answered Feb 03 '11 at 08:17

Kissaki

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score 8 · Answer 2 · answered Feb 03 '11 at 08:25

8

If a thread owns the lock on this, it will go into other synchronized methods/block like hot knife in butter.

answered Feb 03 '11 at 08:25

fastcodejava

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score 3 · Answer 3 · answered Feb 03 '11 at 08:21

In a synchronized block you can re-aquire a lock that you already own. The lock provider (this) will be acquired by the first thread to enter doSomething(). Then in the doSomethingElse() method it will re-aquire it.

The reason to do this? Well there is nothing to say that there is not some other thread entering doSomethingElse from elsewhere. This will prevent any code from executing the block in doSomething() until the thread in doSomethingElse() has released "this".

Edit BTW it can of course block... and it will block any thread that does not own the synchronization object from executing. However it will not (as posted) deadlock.

score 1 · Answer 4 · answered Feb 03 '11 at 08:17

1

Lock already obtanied by the thread executing doSomething method, therefore this thread can execute doSomethingElse method.

You need two lock object to create a deadlock situation. In your case, there is only one, therefore it is impossible to create a deadlock.

answered Feb 03 '11 at 08:17

Gursel Koca

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Well, it should run correctly.. if it is not, you should post all necessary code. – Gursel Koca Feb 03 '11 at 08:22
He means the thread already has the required lock for the secound method so it won't block itself. – kasten Feb 03 '11 at 08:23
Do you mean the code [// do something else] can run correctly? – Hossein Margani Feb 03 '11 at 08:29
I suppose , there is code block instead of comment. if there is only comment, it is no operation. JVM may not even call doSoemthinhElse method, if there is not any code. – Gursel Koca Feb 03 '11 at 08:54
+1 for the explanation that it needs at least two locks to create a deadlock. – jschreiner Mar 25 '14 at 15:21

score 0 · Answer 5 · answered Feb 03 '11 at 08:23

The synchronized use in both case this so if the synchronized block in doSomeThing run, you allready have the lock so you can execute the doSomeThingElse method.

If your code is so simple, it is equivalent to :

public synchronized void doSomeThing() {
    doSomeThingElse();
}
public synchronized void doSomeThingElse() {
    // do something else
}

score 0 · Answer 6 · answered Feb 03 '11 at 08:24

You have already taken the monitor lock in the first synchronized. You should always ensure that the concurrency design does not make the performance impact huge.

One way to ensure this is by synchronizing only the required statements/code.

Let's say now your code will look something like this.

public void doSomeThing()
 {
  synchronized (this) 
   {             
    synchronized (this) 
     {             
      // do something else         
     }           
   }     
 }

Whereas this is what is required

public void doSomeThing()
     {
       doSomeThingElse();     
     }
public void doSomeThingElse()
{
 synchronized (this) 
         {             
          // do something else         
         }
}

Nested synchronized keyword

6 Answers6

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