Knowing that
Reads and writes are atomic for all variables declared volatile
Question1: Can this be understood
As a quickly testable example that may illustrate the previous answers, this yields always a final count of 8:
import java.util.concurrent.atomic.AtomicInteger;
public class ThreadTest_synchronize {
public static void main(String[] args) {
ThreadTest_synchronize tt = new ThreadTest_synchronize ();
try {
tt.go();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
private void go() throws InterruptedException{
MyRunnable t = new MyRunnable();
Thread myThread_1 = new Thread( t, "t1");
Thread myThread_2 = new Thread( t, "t2");
myThread_1.start();
myThread_2.start();
myThread_1.join();
myThread_2.join();
System.out.println("Processing count="+t.getCount());
}
private class MyRunnable implements Runnable{
private AtomicInteger count=new AtomicInteger(0);
@Override
public void run() {
for(int i=1; i< 5; i++){
doSomething(i);
count.getAndAdd(1);
}
}
public AtomicInteger getCount() {
return this.count;
}
private void doSomething(int i) {
try {
Thread.sleep(i*300);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
while this generally does not:
public class ThreadTest_volatile {
public static void main(String[] args) {
ThreadTest_volatile tt = new ThreadTest_volatile ();
try {
tt.go();
} catch (InterruptedException e) {
e.printStackTrace();
}
}
private void go() throws InterruptedException{
MyRunnable t = new MyRunnable();
Thread myThread_1 = new Thread( t, "t1");
Thread myThread_2 = new Thread( t, "t2");
myThread_1.start();
myThread_2.start();
myThread_1.join();
myThread_2.join();
System.out.println("Processing count="+t.getCount());
}
private class MyRunnable implements Runnable{
private volatile int count = 0;
@Override
public void run() {
for(int i=1; i< 5; i++){
doSomething(i);
count++;
}
}
private int add(int count){
return ++count;
}
public int getCount(){
return count;
}
private void doSomething(int i) {
try {
Thread.sleep(i*300);
} catch (InterruptedException e) {
e.printStackTrace();
}
}
}
}
The volatile only gives you additional visibility guarantees, atomic writes/reads for longs/doubles (otherwise not guaranteed by the JLS, yes) and some memory order guarantees. No synchronization (it is possible though to build synchronization blocks starting with just volatile - Dekker's algorithm )
So no, it does not help you with x++
- that's still a read, inc and write and needs some form of synchronization.
One example of volatile is the famous double-checked locking, where we avoid synchronization most of the time because the ordering guarantees are all we need:
private volatile Helper helper = null;
public Helper getHelper() {
if (helper == null) {
synchronized(this) {
if (helper == null) {
helper = new Helper();
}
}
}
return helper;
}
An example where there's absolutely no synchronization involved, is a simple exit flag, here it's not about ordering guarantees but only about the guaranteed visibility
public volatile boolean exit = false;
public void run() {
while (!exit) doStuff();
// exit when exit set to true
}
If another thread sets exit = true
the other thread doing the while loop is guaranteed to see the update - without volatile it may not.
x++; operation is atomic?
No. This reduces to x = x + 1
. The read of x
is atomic, and the write to x
is atomic, but x = x + 1
as a whole is not atomic.
I wonder under what circumstances (if any) it is possible to see a variable marked volatile and not see any methods of blocks marked synchronized (that attempt to access/ modify the variable)?
Well, there are all kinds of approaches to concurrency that don't use synchronized
. There's a wide variety of other locking utilities in Java, and lock-free algorithms that still require things like volatile
: ConcurrentLinkedQueue
is a specific example, though it makes extensive use of "magical" compareAndSet
atomics.