Java – Threaded State Machines in Java

javamultithreadingstate-machineswitch statementsynchronized

Is there a way of Holding a thread in a State waiting for changes?
I mean wait tll something happend (change var, call method, etc..)
perhaps it needs using Event Listeners, or synchronized objects/methods..

The usual aproach for statemachines like this

statemachine example

That uses a do{..}while(true) loop that could work for singled threaded(and no GUI) application but it CANT'T be used with threads..
(at least you want to use a core for every threaded state machine)

So to avoid that processor consuming task a simple (and ugly) way
is a "periodic checker", I mean adding a Sleep
other idea is defining a synchronized object and using wait instead of thread Sleep

example:

do{
    Switch(state)
    {         
        case STATE_A:
            //..A things                         
            break;

        case STATE_B:
            //..B things                         
            break;

        ...

        case STATE_Z:
            //..Z things                         
            break;
    }
    // Here!  =>   wait()?  Thread.sleep(TIME_CONST)? //Hold and yield
}while(powerOn);

The drawback is adding complexity using synchronized or that anything that happen within TIME_CONST would be invisible

I would like to know other ideas for doing this,
thanks !

Best Answer

Yes, you would be using wait/notify. That is what it is for.

Or maybe your state machine does not need to have its own thread. Other threads could call an update method on the state machine when something interesting happens.

Related Solutions

Java – What are the differences between a HashMap and a Hashtable in Java

There are several differences between HashMap and Hashtable in Java:

Hashtable is synchronized, whereas HashMap is not. This makes HashMap better for non-threaded applications, as unsynchronized Objects typically perform better than synchronized ones.
Hashtable does not allow null keys or values. HashMap allows one null key and any number of null values.
One of HashMap's subclasses is LinkedHashMap, so in the event that you'd want predictable iteration order (which is insertion order by default), you could easily swap out the HashMap for a LinkedHashMap. This wouldn't be as easy if you were using Hashtable.

Since synchronization is not an issue for you, I'd recommend HashMap. If synchronization becomes an issue, you may also look at ConcurrentHashMap.

Java – Is Java “pass-by-reference” or “pass-by-value”

Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.

It goes like this:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    // we pass the object to foo
    foo(aDog);
    // aDog variable is still pointing to the "Max" dog when foo(...) returns
    aDog.getName().equals("Max"); // true
    aDog.getName().equals("Fifi"); // false
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // change d inside of foo() to point to a new Dog instance "Fifi"
    d = new Dog("Fifi");
    d.getName().equals("Fifi"); // true
}

In the example above aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.

Likewise:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    foo(aDog);
    // when foo(...) returns, the name of the dog has been changed to "Fifi"
    aDog.getName().equals("Fifi"); // true
    // but it is still the same dog:
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // this changes the name of d to be "Fifi"
    d.setName("Fifi");
}

In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.

For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.

Best Answer

Related Solutions

Java – What are the differences between a HashMap and a Hashtable in Java

Java – Is Java “pass-by-reference” or “pass-by-value”

Related Topic