سؤال

private volatile static Singleton uniqueInstance

In a singleton when using double lock method for synchronization why is the single instance declared as volatile ? Can I achieve the same functionality without declaring it as volatile ?

هل كانت مفيدة؟

المحلول

Without volatile the code doesn't work correctly with multiple threads.

From Wikipedia's Double-checked locking:

As of J2SE 5.0, this problem has been fixed. The volatile keyword now ensures that multiple threads handle the singleton instance correctly. This new idiom is described in The "Double-Checked Locking is Broken" Declaration:

// Works with acquire/release semantics for volatile
// Broken under Java 1.4 and earlier semantics for volatile
class Foo {
    private volatile Helper helper = null;
    public Helper getHelper() {
        Helper result = helper;
        if (result == null) {
            synchronized(this) {
                result = helper;
                if (result == null) {
                    helper = result = new Helper();
                }
            }
        }
        return result;
    }

    // other functions and members...
}

In general you should avoid double-check locking if possible, as it is difficult to get right and if you get it wrong it can be difficult to find the error. Try this simpler approach instead:

If the helper object is static (one per class loader), an alternative is the initialization on demand holder idiom

// Correct lazy initialization in Java 
@ThreadSafe
class Foo {
    private static class HelperHolder {
       public static Helper helper = new Helper();
    }

    public static Helper getHelper() {
        return HelperHolder.helper;
    }
}

نصائح أخرى

The volatile prevents memory writes from being re-ordered, making it impossible for other threads to read uninitialized fields of your singleton through the singleton's pointer.

Consider this situation: thread A discovers that uniqueInstance == null, locks, confirms that it's still null, and calls singleton's constructor. The constructor makes a write into member XYZ inside Singleton, and returns. Thread A now writes the reference to the newly created singleton into uniqueInstance, and gets ready to release its lock.

Just as thread A gets ready to release its lock, thread B comes along, and discovers that uniqueInstance is not null. Thread B accesses uniqueInstance.XYZ thinking that it has been initialized, but because the CPU has reordered writes, the data that thread A has written into XYZ has not been made visible to thread B. Therefore, thread B sees an incorrect value inside XYZ, which is wrong.

When you mark uniqueInstance volatile, a memory barrier is inserted. All writes initiated prior to that of uniqueInstance will be completed before the uniqueInstance is modified, preventing the reordering situation described above.

To avoid using double locking, or volatile I use the follow

enum Singleton {
     INSTANCE;
}

Creating the instance is simple, lazy loaded and thread safe.

Write to a volatile field will happen before any read operation. Below is an example code for better understanding:

private static volatile ResourceService resourceInstance;
//lazy Initialiaztion
public static ResourceService getInstance () {
    if (resourceInstance == null) { // first check
        synchronized(ResourceService.class) {
            if (resourceInstance == null) { // double check
                // creating instance of ResourceService for only one time
                resourceInstance = new ResourceService ();                    
            }
        }
    }
    return resourceInstance;
}

This link can serve you better http://javarevisited.blogspot.com/2011/06/volatile-keyword-java-example-tutorial.html

You can use the follow code:

private static Singleton uniqueInstance;

public static synchronized Singleton getInstance(){
    if(uniqueInstance == null){
        uniqueInstance = new Singleton();
    }
    return uniqueInstance
}
مرخصة بموجب: CC-BY-SA مع الإسناد
لا تنتمي إلى StackOverflow
scroll top