Immutable Objects in multi threaded application - how does it work?

Question

I have this code will work in multithreaded application. I know that immutable object is thread safe because its state cannot be changed. And if we have volatile reference, if is changed with e.g. MyImmutableObject state = MyImmutableObject.newInstance(oldState, newArgs); i.e. if a thread wants to update the state it must create new immutable object initializing it with the old state and some new state arguments) and this will be visible to all other threads. But the question is, if a thread2 starts long operation with the state, in the middle of which thread1 updates the state with new instance, what will happen? Thread2 will use reference to the old object state i.e. it will use inconsistent state? Or thread2 will see the change made by thread1 because the reference to state is volatile, and in this case thread1 can use in the first part of its long operation the old state and in the second part the new state, which is incorrect?

State state = cache.get(); //t1 
Result result1 = DoSomethingWithState(state); //t1 
        State state = cache.get(); //t2
    ->longOperation1(state); //t1
        Result result2 = DoSomethingWithState(state); //t2
             ->longOperation1(state); //t2
   ->longOperation2(state);//t1
cache.update(result1); //t1 
             ->longOperation2(state);//t2
        cache.update(result2);//t2

Result DoSomethingWithState(State state) {
    longOperation1(state);
    //Imaging Thread1 finish here and update state, when Thread2 is going to execute next method
    longOperation2(state);
return result;
}

class cache {
     private volatile State state = State.newInstance(null, null);

    update(result) {
        this.state = State.newInstance(result.getState, result.getNewFactors);

    get(){
       return state;
    }

 }

Answer 1

But the reference is volatile, isn't it making visible the new object state … to the other threads?

No. While a write to a volatile field happens-before every subsequent read of that field, the other thread must re-read that field to get the new value.

Answer 2

What you describe is not thread safe, despite the fact that you are using immutable objects. thread2 continues to act on the object it originally obtained a reference to, even if thread1 assigns a new value of state concurrently. In java, object references are passed by value. So when the state argument to DoSomethingWithState() or longOperation1() gets passed, the method sees only the object reference that was passed, not the changing values later. If you want this kind of code to be thread safe, you need to synchronize all methods that act on the singleton state object.