Hi
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?
Therad1: Thead2: ---------------------------------------------------------------------------------------------- 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); cache.update(result) { this.state = State.newInstance(result.getState, result.getNewFactors); get(){ return state; } } }