On Thu, 30 Jun 2022 18:24:14 GMT, Xue-Lei Andrew Fan <[email protected]> wrote:
>> test/lib/jdk/test/lib/util/ForceGC.java line 70:
>>
>>> 68: // But it is fine. For most cases, the 1st GC is
>>> sufficient
>>> 69: // to trigger and complete the cleanup.
>>> 70: queue.remove(200L);
>>
>> If `remove()` returns a non-null value, then it is safe to break out of the
>> loop.
>> The GC has cleared the ref. And the final `getAsBoolean()` below will return
>> the condition.
>
> I'm not sure if all unused object will be collected in one GC call always.
> The gc() specification says "When control returns from the method call, the
> Java Virtual Machine has made a best effort to reclaim space from all unused
> objects. ... There is also no guarantee that this effort will determine the
> change of reachability in any particular number of objects, or that any
> particular number of {@link java.lang.ref.Reference Reference} objects will
> be cleared and enqueued." But from the spec, I did not get a clear answer
> for the question.
>
> If the `booleanSupplier` object could be cleared in a collection other than
> the `ref` collection, the current code may be safer.
True, knowing when GC is 'done' is not deterministic except for a specify
Reference to a specific object.
System.gc is just a request, the checking for an object can more quickly exit
the loop.
The code is as is, and already commented, might wait an extra cycle, but only
in the case of a race between the requested predicate and the object being
reclaimed. Ok as it.
-------------
PR: https://git.openjdk.org/jdk/pull/8979
