What Volatile Means in Java

Today, I'm going to talk about what volatile means in Java. I've sort-of covered this in other posts, such as my posting on the ++ operator, my post on double-checked locking and the like, but I've never really addressed it directly.

First, you have to understand a little something about the Java memory model. I've struggled a bit over the years to explain it briefly and well. As of today, the best way I can think of to describe it is if you imagine it this way:

• Each thread in Java takes place in a separate memory space (this is clearly untrue, so bear with me on this one).


• You need to use special mechanisms to guarantee that communication happens between these threads, as you would on a message passing system.


• Memory writes that happen in one thread can "leak through" and be seen by another thread, but this is by no means guaranteed. Without explicit communication, you can't guarantee which writes get seen by other threads, or even the order in which they get seen.

The Java volatile modifier is an example of a special mechanism to guarantee that communication happens between threads. When one thread writes to a volatile variable, and another thread sees that write, the first thread is telling the second about all of the contents of memory up until it performed the write to that volatile variable.

At this point, I usually rely on a visual aid, which we call the "two cones" diagram, but which my officemate insists on calling the "two trapezoids" diagram, because he is picky. ready is a volatile boolean variable initialized to false, and answer is a non-volatile int variable initialized to 0.




The first thread writes to ready, which is going to be the sender side of the communications. The second thread reads from ready and sees the value the first thread wrote to it. It therefore becomes a receiver. Because this communication occurs, all of the memory contents seen by Thread 1, before it wrote to ready, must be visible to Thread 2, after it reads the value true for ready.

This guarantees that Thread 2 will print "42", if it prints anything at all.

If ready were not volatile, what would happen? Well, there wouldn't be anything explicitly communicating the values known by Thread 1 to Thread 2. As I pointed out before, the value written to the (now non-volatile) ready could "leak through" to Thread 2, so Thread 2 might see ready as true. However, the value for answer might not leak through. If the value for ready does leak through, and the value for answer doesn't leak through, then this execution will print out 0.

We call the communications points "happens-before" relationships, in the language of the Java memory model.

(Minor niggle: The read of ready doesn't just ensure that Thread 2 sees the contents of memory of Thread 1 up until it wrote to ready, it also ensures that Thread 2 sees the contents of memory of any other thread that wrote to ready up until that point.)

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With this in mind, let's look at the Double-Checked Locking example again. To refresh your memory, it goes like this:

class Foo {
  private volatile Helper helper = null;
  public Helper getHelper() {
    if (helper == null) {
      synchronized(this) {
        if (helper == null) {
          helper = new Helper();
        }
      }
    }
  return helper;
}

The object of the double-checked locking pattern is to avoid synchronization when reading a lazily constructed singleton that is shared between threads. If you have already constructed the object, the helper field will not be null, so you won't have to perform the synchronization.

However, this is only part of the solution. If one thread creates the object, it has to communicate the contents of its memory to another thread. Otherwise, the object will just sit in the first thread's memory. How do we communicate the contents of memory to another thread? Well, we can use volatile variables. That's why helper has to be volatile -- so that other threads see the fully constructed object.

Locking in Java also forms these "happens-before" communication points. An unlock is the sender side, and a lock on the same variable is the receiver side. The reason that doesn't work for (non-volatile) double-checked locking is that only the writing thread ever performs the locking. The whole point of the idiom is that the reader side doesn't do the locking. Without the explicit communication in the form of the volatile variable, the reading thread will never see the update performed by the writer thread.

Let me know if that makes sense.