Java 中的惰性布尔关闭标志是否需要 volatile？

Question

Assume the following code

public class Singleton {

  boolean shuttingDown = false;


  void action() {
    if (shuttingDown) {
      throw new RuntimeException("already shutting down");
    } 
    // do some more stuff
  }

  // Called by a single thread only
  void onShutDown() {
    shuttingDown = true;
    // perform some more actions to remedy the class
  }
}

Basically I want to block all upcoming calls to action() with an exception. I know that setting shuttingDown is an atomic operation. However the question is if I would need to make shuttingDown volatile to make the change visible to the other threads, which might be re-used from a thread pool.

I saw this oracle tutorial and also the javadoc on AtomicBoolean . However the latter uses volatile , too. I set the value only once in a single thread, so I don't need the locking mechanisms provided by AtomicBoolean. I just want to make the change visible to all threads as soon as the variable is updated.

As far as I understand the oracle tutorial, the update operation is atomic , that is no other thread can intecept while setting the value. The question is, when will the updated value be populated to the other threads (if at all?).

Answer 1

The short answer is yes, you must make it volatile.

Although the operation is atomic, for it to be guaranteed visible to other threads you need to establish a happens before relationship. Volatile is the simplest way to do this.