RFR: 8257831: Suspend with handshakes [v4]
Richard Reingruber
rrich at openjdk.java.net
Sat Apr 10 07:41:25 UTC 2021
On Fri, 9 Apr 2021 16:12:16 GMT, Daniel D. Daugherty <dcubed at openjdk.org> wrote:
>> src/hotspot/share/runtime/handshake.cpp line 630:
>>
>>> 628: // exiting.
>>> 629: }
>>> 630: }
>>
>> I need a little help learning the steps of this dance :)
>>
>> We reach here in _thread_in_vm. We cannot be suspended in this state. There
>> might be another thread waiting to handshake us to suspend us but why can't we
>> just ignore that and do the `_handshakee->set_exiting();` even without locking
>> HandshakeState::_lock?
>>
>> Adding a handshake operation is lock free, so even if the check
>> `SafepointMechanism::should_process(_handshakee)` in L619 returns false a new
>> operation to process could have been added concurrently such that
>> `SafepointMechanism::should_process(_handshakee)` would return true when we
>> execute `_handshakee->set_exiting();` in L620. What am I missing?
>
> `HandshakeState::suspend()` is a synchronous handshake and adding the
> handshake to the queue is lock free, but the execution of the synchronous
> handshake itself requires a `HandshakeState::claim_handshake()` call which
> does acquire the lock in question. We (the suspend requester) hold the lock
> while the handshake is being processed so we either detect that
> _handshakee->set_exiting() won the race (in the target thread) or we (the
> suspend requester) win the race of setting the suspend flag so the target
> thread can't exit yet.
>
> Hopefully that helps explain this dance.
Hi Dan,
thanks for picking up my question!
> `HandshakeState::suspend()` is a synchronous handshake and adding the
> handshake to the queue is lock free, but the execution of the synchronous
> handshake itself requires a `HandshakeState::claim_handshake()` call which
> does acquire the lock in question.
My point would be that the attempt to execute the synchronous handshake for
suspending a thread that is just about to call HandshakeState::thread_exit()
cannot make progress (blocks) while the target thread is not safe
(_thread_in_vm).
A synchronous handshake requires the target thread to be in a safe state for the
requester to execute the handshake operation. When executing
HandshakeState::thread_exit() the suspendee is _thread_in_vm. And the requester
will find it to be `_not_safe` when calling `possibly_can_process_handshake()`
before calling `HandshakeState::claim_handshake()` or when calling
`can_process_handshake()` afterwards. In both cases try_process() returns with
failure _not_safe and the lock is not held.
++
546 if (!possibly_can_process_handshake()) {
547 // JT is observed in an unsafe state, it must notice the handshake itself
548 return HandshakeState::_not_safe;
549 }
550
551 // Claim the mutex if there still an operation to be executed.
552 if (!claim_handshake()) {
553 return HandshakeState::_claim_failed;
554 }
555
556 // If we own the mutex at this point and while owning the mutex we
557 // can observe a safe state the thread cannot possibly continue without
558 // getting caught by the mutex.
559 if (!can_process_handshake()) {
560 _lock.unlock();
561 return HandshakeState::_not_safe;
562 }
So isn't being unsafe sufficient to sync with suspend requests?
-------------
PR: https://git.openjdk.java.net/jdk/pull/3191
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