Volatile in Java

If, in the following example, one thread repeatedly calls the method one() and another thread repeatedly calls the method two():

class Test {
static int i = 0, j = 0;
static void one() { i++; j++; }
static void two() {
System.out.println("i=" + i + " j=" + j);
}
}
then method two could occasionally print a value for j that is greater than the value of i, because the example includes no synchronization and the shared values of i and j might be updated out of order.

One way to prevent this out-of-order behavior would be to declare methods one and two to be synchronized.

Another approach would be to declare i and j to be volatile:

class Test {
static volatile int i = 0, j = 0;
static void one() { i++; j++; }
static void two() {
System.out.println("i=" + i + " j=" + j);
}
}
This allows method one and method two to be executed concurrently, but guarantees that accesses to the shared values for i and j occur exactly as many times, and in exactly the same order, as they appear to occur during execution of the program text by each thread. Therefore, the shared value for j is never greater than that for i, because each update to i must be reflected in the shared value for i before the update to j occurs