Java: How to do double-checked-locking with an array element?

Question

This is what my code currently looks like:

private boolean[] isInitialized = new boolean[COUNT];

private void ensureInitialized(int i) {
    if (! isInitialized[i]) {
        initialize(i);
        isInitialized[i] = true;
    }
}

Now I want to have it thread-safe. I know that double-checked-locking in Java is "teh 3vilness!!1", but since ensureInitialized may be called very often, I don't want it to be synchronized. So I am thinking of doing this:

private boolean[] isInitialized = new boolean[COUNT];

private void ensureInitialized(int i) {
    if (! isInitialized[i]) {
        synchronized (this) {
            if (! isInitialized[i]) {
                initialize(i);
                isInitialized[i] = true;
            }
        }
    }
}

Now what do I have to do to make this actually thread safe?
Some subquestions:

Making isInitialized volatile is not necessary, since the variable is not changed, right?
The array elements are changed, how can I make those volatile?
Are there generally better ways to do this?

(Also notice that this is an instance method, so static initializer will no work)

mindas · Accepted Answer · 2011-01-25 22:04:38Z

3

You could also consider using Suppliers.memoize(Supplier<T> delegate) from guava library.

answered Jan 25, 2011 at 22:04

mindas

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2 Comments

Guillaume Over a year ago

I like this kind of answer: double checked locking is such an atrocity ! I don't want to see any bit of it im my code :)

mindas Over a year ago

I second that. After I found this idiom I have never looked back. The advantage of this idiom is that library guys will always try to gravitate towards the best approach and if they find one that is better, they will change it and you don't have to rewrite your thousand double-checked blocks across your code base.

Dewfy · Accepted Answer · 2010-10-09 18:05:50Z

1

Please note that java implementation of double-checking is called "broken pattern" because it was proven as fail (for example see http://www.cs.umd.edu/~pugh/java/memoryModel/DoubleCheckedLocking.html). To work around just use atomic operation. Follow is sample how to build thread safe singleton:

static AtomicReferenceArray<Boolean> instance = 
   new AtomicReferenceArray<Boolean>(COUNT);

private void ensure(int i)
{
    if(!instance.get(i) )
    {
        synchronized(this){
        if( !instance.get(i) ){
           initialize(i);
           instance.set(i, true);
        }
    }
}

answered Oct 9, 2010 at 18:05

Dewfy

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6 Comments

President James K. Polk Over a year ago

Is there any evidence that java's synchronization is unnecessarily slow? In other words, yes, it is slow when there is contention, but when there is no contention it's fast.

Bozho Over a year ago

+1, but an explanation is due. The double-checking lock is not broken since Java 1.5 - using the volatile keyword makes it work fine. However, as the OP rightly assumed, the reads and writes to elements of the array are not volatile. What AtomicXArray classes do is do a volatile read for each element (See the source - unsafe.getObjectVolatile()).

Tim Bender Over a year ago

-1, Double check locking is not broken. The rest of the answer is technically correct. Maybe less preaching and more technical solutions?

Dewfy Over a year ago

@Tim Bender - just go to link and/or read comment of @Bozho. Or at least try to google around phrase "java double checking" "broken pattern"

Dewfy Over a year ago

@GregS - it is rather funny, why we should use synchronization at all if no "contention"? When you sure that singleton used only by single thread lazy-load pattern is enough.

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