]
Vladimir Blagojevic reassigned ISPN-799:
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Assignee: Vladimir Blagojevic (was: Manik Surtani)
JoinTask as it invalidates L1 entries should be given precedence in
acquiring locks
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Key: ISPN-799
URL:
https://jira.jboss.org/browse/ISPN-799
Project: Infinispan
Issue Type: Bug
Components: Locking and Concurrency
Affects Versions: 4.2.0.CR1
Reporter: Vladimir Blagojevic
Assignee: Vladimir Blagojevic
Fix For: 4.2.0.Final, 5.0.0.Final
The SingleJoinTest transaction test failure itself is intermittent due to the way
addresses are organised in the hash wheel, so you are correct that it is a timing issue.
Anyway, it still is a very real problem. Just to re-iterate and to make sure we are
talking about the same thing:
1. View is {A, B, C}
2. K is mapped to {A, B}
3. A tx starts to update K, and is prepared. Locks now held for K on {A, B}
4. D joins. D is placed on the hash wheel between A and B. So the new view is {A, D,
B, C}
5. As per the test (artificial, I know, but could still happen), the tx waits for a long
time before committing. In the case of the test, artificially waits until D has finished
joining before committing, by use of a latch.
6. D never joins as even though it receives the prepare for the tx and could potentially
commit itself (as a new owner), it fails as it is unable to invalidate K on B.
There are a few solutions here:
1) This is pretty easy to detect. Attempt to acquire the lock with a smaller lock
acquisition timeout and if the transaction is still stuck, abort the transaction and
proceed with the join.
2) If the blocking node is *not* the transaction originator (as in this case: the tx was
started on A), then just force lock removal and tx rollback on B *only*. Let the tx
complete on A, since the new joiner will receive the transactional event and will be able
to apply it as a new owner.
My vote is to go for solution 1 - a bit more crude, but 2 would be very complex to
implement. And even then, would only solve for the invalidation being blocked on a node
that did not originate the transaction. E.g., the tx originated on A but the lock issue
was on B. If, however, the tx originated on B, *and* B no longer owns the entry in
question, then 2 is no longer a solution and the only solution would be 1.
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