java synchronization gone wild and back

[juan_gandhi]

Hey, so, you, like everyone else, decided to make your static initialization thread-safe. You have a singleton that is hard to instantiate, and You have something like this:

static My worker;

static My worker() {
  if (worker == null) {
    worker = new MyImplementation();
  }
  return worker;
}

and it does not work, because, say, a dozen of servlet threads in parallel are trying to reinitialize your singleton like crazy; a minute that it takes to instantiate one, turns into 12 minutes, since all 12 are trying to do the same.

Okay, you synchronize it:

static My worker;

static synchronized worker() {
  if (worker == null) {
    worker = new MyImplementation();
  }
  return worker;
}

In this case you lock execution on the whole class; your class may have a bunch of other singletons, and they will all stand in line waiting for this specific singleton to be retrieved - looks stupid, right? so you introduce a gentle check:

static My worker;

static worker() {
  if (worker == null) {
    synchronized {
      worker = new MyImplementation();
    }
  }
  return worker;
}

See, before instantiating, you check whether it is instantiated already.

The trouble is, your twelve threads will get through this check and wait on the same barrier before synchronized block, and then will instantiate the singleton, one after another, taking 12 minutes in total. Okay, you get angry, you open a book and steal a solution, the famous double-check trick:

static My worker;

static worker() {
  if (worker == null) {
    synchronized {
      if (worker == null) {
        worker = new MyImplementation();
      }
    }
  }
  return worker;
}

Now you feel good, the threads would not reinitialize the same singleton, because when the second threads penetrates the synchronized block, worker is already not null. You feel good... but then you open Josh and Neal's Java Puzzlers, and see that oops, this won't work on a multiprocessor machine. Why? See, by the time the first thread leaves the syncrhonized block, the value of worker may not reach the memory shared by the threads.

So, the solution suggested is this:

static My worker;
private interface HR {
  My getWorker();
}

private static HR catbert = new HR() {
  My worker = new MyImplementation();
  My getWorker() {
    return worker;
  }
}

static worker() {
  return catbert.getWorker();
}

See, the trick is that the singleton is provided by a static class, which is loaded strictly once, which is guaranteed by the class loader.

Now enters "new Java 5", not sure which one, 1.5.0.6 or 1.5.0.7. Bug fixed. All you have to do is declare My worker volatile:

volatile static My worker;

static worker() {
  if (worker == null) {
    synchronized {
      if (worker == null) {
        worker = new MyImplementation();
      }
    }
  }
  return worker;
}

So, now we are back to circle 3: after a bugfix, double checks are legal again.

Comments

[ex-ex-zhuzh.livejournal.com]

если весь доступ к переменной внутри pthread_mutex_lock(), то проблем нет. так обычно и бывает. но у нас запись внутри, а чтение-то снаружи. это не работает.

[birdwatcher.livejournal.com]

Так для того и читаем второй раз внутри, казалось бы?

[ex-ex-zhuzh.livejournal.com]

когда читаем внутри, нет проблем. проблема, когда НЕ читаем, т.е. там не NULL. то есть указатель на объект уже записан в память, а состояние самого объекта ещё нет. на унипроцессоре это нормально, система гарантирует, что когда состояние понадобится, оно уже будет в памяти (система знает, когда именно оно понадобится). на мультипроцессоре это не так.

[birdwatcher.livejournal.com]

Разве, когда выполняется эта строчка,
   worker = new MyImplementation();
все остальные процессоры не стоят? Потому что если они стоят, то worker либо null, либо показывает на нормальный объект.

[ex-ex-zhuzh.livejournal.com]

не стоят, нет, с чего бы?

[birdwatcher.livejournal.com]

Ага, увидел. Спасибо.