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Home/ Questions/Q 8734191
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T09:47:32+00:00 2026-06-13T09:47:32+00:00

The following code snippet is taken from android JellyBean ReferenceQueue.java in the libcore project.

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The following code snippet is taken from android JellyBean ReferenceQueue.java in the libcore project.

Can someone tell me why a synchronized block is used, synchronizing on ReferenceQueue.class, instead of adding the synchronized qualifier to the method?
Are these two approaches functionally equivalent in this instance?

From similar questions that I looked at it seems to be more efficient to make the method synchronized.

Cheers,
Matt

public class ReferenceQueue<T> {
... <snip> ...
public static Reference unenqueued = null;

static void add(Reference<?> list) {
    synchronized (ReferenceQueue.class) {
        if (unenqueued == null) {
            unenqueued = list;
        } else {
            Reference<?> next = unenqueued.pendingNext;
            unenqueued.pendingNext = list.pendingNext;
            list.pendingNext = next;
        }
        ReferenceQueue.class.notifyAll();
    }
}
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  1. Editorial Team
    Editorial Team
    2026-06-13T09:47:33+00:00Added an answer on June 13, 2026 at 9:47 am

    They are exactly equivalent except for the method signature. When you make a static method synchronized it is the same as synchronizing the full body of the method on the class token. When you make a non static method synchronized it is the same as synchronizing on the this pointer. The method signature difference is rarely relevant but for instance it can be a compiler warning to override a synchronized method and make the overriding method unsychronized.

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