Why is volatile keyword not allowed for local variables?

问题

Consider the snippet:

If in a main thread, I have this inside a method-

volatile CountDownLatch latch = new CountDownLatch(3);

new Thread(new ProcessThread("Worker1",latch, 20000)).start();//20 secs
new Thread(new ProcessThread("Worker2",latch, 60000)).start();//60 secs
new Thread(new ProcessThread("Worker3",latch, 40000)).start();//40 secs

I see that volatile is shown as an illegal modifier. And only final is permitted. And final guarantees initialization safety.

public static class ProcessThread implements Runnable {
  final CountDownLatch latch;
  final long workDuration;
  final String name;

  public ProcessThread(String name, CountDownLatch latch, long duration){
      this.name= name;
      this.latch = latch;
      this.workDuration = duration;
  }
}

The object below i.e new CountDownLatch(3) is properly constructed but I also want to make sure that the reference latch to which the above object is assigned is guaranteed to be visible to the code below it.

final CountDownLatch latch = new CountDownLatch(3);

Does the above code guarantee initialization so that latch is perfectly visible to the code below i.e

new Thread(new ProcessThread("Worker1",latch, 20000)).start();

回答1:

And final guarantees initialization safety.

Not on local variables: it just stops you reassigning that variable.

final CountDownLatch latch = new CountDownLatch(3);

Does the above code will guarantee initialization so that latch is perfectly visible to the code below i.e

No. It's this code that guarantees it:

public static class ProcessThread implements Runnable {

    final CountDownLatch latch;

    // Plus the assignment in the constructor.

}

final fields are guaranteed to be visible once the constructor completes (normally). From JLS Sec 17.5:

An object is considered to be completely initialized when its constructor finishes. A thread that can only see a reference to an object after that object has been completely initialized is guaranteed to see the correctly initialized values for that object's final fields.

回答2:

Local variables live on the stack; and of course, when you invoke the same method twice, they have all their local variables on their individual stacks.

volatile only makes sense when multiple threads would be writing to the same memory location (on the heap).

Which makes absolutely no sense for local variables from within the body of a method!

回答3:

Operations you perform locally will not have visibility or interference issues by other threads so it does not make sense to declare local variable volatile.

来源：https://stackoverflow.com/questions/39952937/why-is-volatile-keyword-not-allowed-for-local-variables