Question

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 ?

Answer 1

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;
    }
}