RFR: 8355726: LinkedBlockingDeque fixes and improvements

[kabutz]

On Thu, 8 May 2025 08:59:59 GMT, Viktor Klang <vklang at openjdk.org> wrote:

>> We logged several bugs on the LinkedBlockingDeque. This aggregates them into a single bug report and PR.
>> 
>> 1. LinkedBlockingDeque does not immediately throw InterruptedException in put/take
>> 
>> The LinkedBlockingDeque does not behave consistently with other concurrency components. If we call putFirst(), putLast(), takeFirst(), or takeLast() with a thread that is interrupted, it does not immediately throw an InterruptedException, the way that ArrayBlockingQueue and LInkedBlockingQueue does, because instead of lockInterruptibly(), we call lock(). It will only throw an InterruptedException if the queue is full (on put) or empty (on take). Since interruptions are frequently used as a shutdown mechanism, this might prevent code from ever shutting down.
>> 
>> 2. LinkedBlockingDeque.clear() should preserve weakly-consistent iterators
>> 
>> LinkedBlockingDeque.clear() should preserve weakly-consistent iterators by linking f.prev and f.next back to f, allowing the iterators to continue from the first or last respectively. This would be consistent with how the other node-based weakly consistent queues LinkedBlockingQueue LinkedTransferQueue, ConcurrentLinkedQueue/Deque work.
>> 
>> The LBD already supports self-linking, since that is done by the unlinkFirst() and unlinkLast() methods, and the iterators and spliterator thus all support self-linking.
>> 
>> This can be fixed very easily by linking both f.prev and f.next back to f.
>> 
>> 3. LinkedBlockingDeque offer() creates nodes even if capacity has been reached
>> 
>> In the JavaDoc of LinkedBlockingDeque, it states: "Linked nodes are dynamically created upon each insertion unless this would bring the deque above capacity." However, in the current implementation, nodes are always created, even if the deque is full. This is because count is non-volatile, and we only check inside the linkFirst/Last() methods whether the queue is full. At this point we have already locked and have created the Node. Instead, the count could be volatile, and we could check before locking.
>> 
>> In the current version, calling offer() on a full LinkedBlockingDeque creates unnecessary objects and contention. Similarly for poll() and peek(), we could exit prior to locking by checking the count field.
>> 
>> 4. LinkedBlockingDeque allows us to overflow size with addAll()
>> 
>> In LinkedBlockingDeque.addAll() we first build up the chain of nodes and then add that chain in bulk to the existing nodes. We count the nodes in "int n" and then whilst hol...
>
> src/java.base/share/classes/java/util/concurrent/LinkedBlockingDeque.java line 865:
> 
>> 863:         long n = 0;
>> 864:         for (E e : c) {
>> 865:             Objects.requireNonNull(e);
> 
> This makes me wonder: Does it make sense to create new nodes if we don't track if they will still fit into the capacity?

We could if you like, but that would subtly change the current behaviour. I tried to make as few changes as possible.

> Out of curiosity, how does `it.remove()` work under these conditions?

If we call it.remove() on the first element, it delegates to unlinkFirst() (if we are using an ascending iterator), and unlinkLast (if we are using a descending iterator). Similarly, if we call it.remove() on the last element it will call unlinkLast() or unlinkFirst(). With unlinkFirst(), it will make f.next = f (thus linking back to itself) and with unlinkLast(), it will make l.prev = l.

If we call it.remove() on a middle element, then we simply link the p.next = n; n.prev = p; and does not do self-linking. Thus if we have an LBD with 1,2,3,4,5 with two iterators pointing onto 3, if one of them removes it, then the other will continue with 3 (cached), 4, 5, and it won't go back to the beginning and see duplicate elements.

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PR Review Comment: https://git.openjdk.org/jdk/pull/24925#discussion_r2079836006
PR Review Comment: https://git.openjdk.org/jdk/pull/24925#discussion_r2079833797