Static variables accessed in an Instance Method

[Denise Advincula]

This is from Enthuware (QID: com.enthuware.ets.scjp.v6.2.388).

Consider the following code:
 
class Test extends Thread {
static int x, y;
public synchronized void run() {
for ( ;; )  {
x++; y++; System.out.println(x+" "+y); 
}
}

public static void main(String [] args) {
new Test().start();
new Test().start();
}
}
 
What will the above code print?
 
a. It will keep on printin same values for x and y incrementing by 1 on each line.
 
b. It will keep on printin same values for x and y but they may be incrementing by more than 1 on each line.
 
c. It may print different values for x and y but both the values will be incrementing by 1 on each line.
 
d. You cannot say anything about the values.

I answered d because I thought that the code won't even compile because the instance method run is accessing the static variables x and y. My answer is correct but my thoughts didn't match the explanation:

You may be tempted by the synchronized keyword on the run method. But note that, there are two diferent thread objects and both the threads are accquiring locks for their own thread object. So, in this case synchronized doesn't make any sense.

I tried to compile and run the program, and it worked. I'm confused here, how come the static variables x and y were accessed successfully in the run method, which is an instance method? Or am I just seeing it the wrong way?
[ September 08, 2008: Message edited by: Denise Saulon ]

[Seetharaman Venkatasamy]

-------------------
for(; ; ) {}
-------------------

how many time it will run?

Hope This Helps
[ September 08, 2008: Message edited by: seetharaman venkatasamy ]

[Denise Advincula]

Hi, thank you for your reply.

Yes it's an infinite loop. But should this program compile first before being able to run? Sorry but these static variables are confusing me.

[Ananth Chellathurai]

Dear Denise,

If suppose if the for loop is like

for ( int i =0; i < 2; i++ )

. Then you will get

1 1
2 2
3 3
4 4

printed as both x and y are static and accessed from two different threads both threads will use the same reference of x and y, and when x and y gets incremented by 1 from thread 1 the thread 2 will get x+1 when it access x.

x and y are members of class Test and run belongs to the same class. What makes you think x and y can not be accessed by run method?

Ananth Chellathurai
[ September 08, 2008: Message edited by: Ananth Chellathurai ]

[Henry Wong]

I tried to compile and run the program, and it worked. I'm confused here, how come the static variables x and y were accessed successfully in the run method, which is an instance method? Or am I just seeing it the wrong way?

I think you may be "remembering" what you learned the wrong way. It is perfectly fine for non-static methods to access static variables. Java doesn't let you access instance variable from static methods, not the other way around.

Henry

[Henry Wong]

The reason the answer is "d. You cannot say anything about the values" is due to threading. The synchronization on the run() method doesn't really do anything. They are separate instances, hence, separate locks.

So, a race condition exist between the two threads that access the static variables. And since increment operations are not atomic, the increment may sometimes not work correctly -- hence, producing values that "you cannot say anything about".

Henry

[Denise Advincula]

Originally posted by Henry Wong:


I think you may be "remembering" what you learned the wrong way. It is perfectly fine for non-static methods to access static variables. Java doesn't let you access instance variable from static methods, not the other way around.

Henry

Thank you guys. Actually it's really the static variables that are confusing me. Yes, I must have remembered it the wrong way. I'm trying to search K&B about static variables being accessed in an instance method. I couldn't find it yet but thanks again for informing me anyway.

[Lok Manick]

static variables can even be accessed with the reference of the class.
ex: objref.staticVariable = value; //valid
Only think that you need to remember is that all the references of the object will access the same static variable.