Khalid Mughal - Threads Page 282

[Shah Chunky]

Hello all...
Khalid Mughal - Threads Page 282
public class VolThread extends Thread
{
static Object lock1 = new Object();
static Object lock2 = new Object();
static volatile int i1,i2,j1,j2,k1,k2;
public void run()
{
while (true)
{
doit();
check();
}
}
void doit()
{
synchronized (lock1)
{
i1++;
}
j1++;
synchronized (lock2)
{
k1++;
k2++;
}
j2++;
synchronized (lock1)
{
i2++;
}
}
void check()
{
if (i1!=i2) System.out.println("i");
if (j1!=j2) System.out.println("j");
if (k1!=k2) System.out.println("k");
}
public static void main(String[] args)
{
System.out.println("Hello ");
new VolThread().start();
new VolThread().start();
}
}
For the above Program, I have 2 Questions :-
Q.1) Does removing the Word Volatile in above have any
effect on the Outcome ?
Q. 2) Suppose if Thread 1 is Executing Synchronized(lock1), then Thread 2
can Execute the statement j1++. OR Thread 2 has to wait for Thread1
to Release the lock ?
Thanks.

[vivek nashine]

Hello friend,
The answers for U'r queries:
1)Yes,not declaring the variables in this code can have a changeable effect on the outcome,b'coz in this code there is a possibility of two thread accessing the same variable.
2)no, any other step after synchronized will be executed only after the current syncronized block's execution is completed by the thread.

[Scott Appleton]

My understanding is that Thread2 *can* execute the j1++ line while Thread1 holds the monitor on lock1 because the j1++ line of code is not within the synchronized block of code where the monitor is applicable.

[Mikael Jonasson]

Remeber that there is no end to the while-loop, and there is no telling when a given thread will loose focus. Thread 2 might be right after the syncronzie block when it stops, then thread 1 just enters it when it stops. This will give a situation where thread 1 is in the sync-block, while thread 2 executes j1++;
/Mike

[jeena jose]

can someone explain to me the importance of volatile keyword in detail???