有关单例类和线程的问题

Question

I'm trying to learn about singleton classes and how they can be used in an application to keep it thread safe. Let's suppose you have an singleton class called IndexUpdater whose reference is obtained as follows:

 public static synchronized IndexUpdater getIndexUpdater() {
    if (ref == null)
        // it's ok, we can call this constructor
        ref = new IndexUpdater();
    return ref;
}

private static IndexUpdater ref;

Let's suppose there are other methods in the class that do the actual work (update indicies, etc.). What I'm trying to understand is how accessing and using the singleton would work with two threads. Let's suppose in time 1, thread 1 gets a reference to the class, through a call like this IndexUpdater iu = IndexUpdater.getIndexUpdater(); Then, in time 2, using reference iu, a method within the class is called iu.updateIndex by thread 1. What would happen in time 2, a second thread tries to get a reference to the class. Could it do this and also access methods within the singleton or would it be prevented as long as the first thread has an active reference to the class. I'm assuming the latter (or else how would this work?) but I'd like to make sure before I implement.

Thank you,

Elliott

Answer 1

Your assumption is wrong. Synchronizing getIndexUpdater() only prevents more than one instance being created by different threads calling getIndexUpdater() at (almost) the same time.

Without synchronization the following could happen: Thread one calls getIndexUpdater(). ref is null. Thread 2 calls getIndexUpdater(). ref is still null. Outcome: ref is instantiated twice.

Answer 2

Since getIndexUpdater() is a synchronized method, it only prevents threads from accessing this method (or any method protected by the same synchronizer) simultaneously. So it could be a problem if other threads are accessing the object's methods at the same time. Just keep in mind that if a thread is running a synchronized method, all other threads trying to run any synchronized methods on the same object are blocked.

More info on: http://download.oracle.com/javase/tutorial/essential/concurrency/syncmeth.html

Answer 3

You are conflating the instantiation of a singleton object with its use. Synchronizing the creation of a singleton object does not guarantee that the singleton class itself is thread-safe. Here is a simple example:

public class UnsafeSingleton {
    private static UnsafeSingleton singletonRef;
    private Queue<Object> objects = new LinkedList<Object>();

    public static synchronized UnsafeSingleton getInstance() {
        if (singletonRef == null) {
            singletonRef = new UnsafeSingleton();
        }

        return singletonRef;
    }

    public void put(Object o) {
        objects.add(o);
    }

    public Object get() {
        return objects.remove(o);
    }
}

Two threads calling getInstance are guaranteed to get the same instance of UnsafeSingleton because synchronizing this method guarantees that singletonRef will only be set once. However, the instance that is returned is not thread safe , because (in this example) LinkedList is not a thread-safe queue. Two threads modifying this queue may result in unexpected behavior. Additional steps have to be taken to ensure that the singleton itself is thread-safe, not just its instantiation. (In this example, the queue implementation could be replaced with a LinkedBlockingQueue , for example, or the get and put methods could be marked synchronized .)

Answer 4

Then, in time 2, using reference iu, a method within the class is called iu.updateIndex by thread 1. What would happen in time 2, a second thread tries to get a reference to the class. Could it do this and also access methods within the singleton ...?

The answer is yes. Your assumption on how references are obtained is wrong. The second thread can obtain a reference to the Singleton. The Singleton pattern is most commonly used as a sort of pseudo-global state. As we all know, global state is generally very difficult to deal with when multiple entities are using it. In order to make your singleton thread safe you will need to use appropriate safety mechanisms such as using atomic wrapper classes like AtomicInteger or AtomicReference (etc...) or using synchronize (or Lock ) to protect critical areas of code from being accessed simultaneously.

Answer 5

The safest is to use the enum-singleton.

public enum Singleton {
  INSTANCE;
  public String method1() {
    ...
  }
  public int method2() {
    ...
  }
}

Thread-safe, serializable, lazy-loaded, etc. Only advantages !

Answer 6

When a second thread tries to invoke getIndexUpdater() method, it will try to obtain a so called lock , created for you when you used synchronized keyword. But since some other thread is already inside the method, it obtained the lock earlier and others (like the second thread) must wait for it.

When the first thread will finish its work, it will release the lock and the second thread will immediately take it and enter the method. To sum up, using synchronized always allows only one thread to enter guarded block - very restrictive access.

Answer 7

The static synchronized guarantees that only one thread can be in this method at once and any other thread attempting to access this method (or any other static synchronized method in this class) will have to wait for it to complete.

IMHO the simplest way to implement a singleton is to have a enum with one value

enum Singleton {
    INSTANCE
}

This is thread safe and only creates the INSTANCE when the class is accessed.

Answer 8

As soon as your synchronized getter method will return the IndexUpdater instance (whether it was just created or already existed doesn't matter), it is free to be called from another thread. You should make sure your IndexUpdater is thread safe so it can be called from multiple threads at a time, or you should create an instance per thread so they won't be shared.