Hi,
The following originated from discussion on a feature in HBase which relies
on zookeeper for coordination.
A task is removed from the tasks map. An async ZK request is pending to
remove the task's zk node.
Another request arrives for this to-be-deleted task. A new in-memory task
structure is
are all of these requests being issued from the same zookeeper handle?
ben
On Thu, Dec 22, 2011 at 6:09 PM, Ted Yu wrote:
> Hi,
> The following originated from discussion on a feature in HBase which relies
> on zookeeper for coordination.
>
> A task is removed from the tasks map. An async ZK req
The requests come from same zhandle.
Cheers
On Thu, Dec 22, 2011 at 6:09 PM, Ted Yu wrote:
> Hi,
> The following originated from discussion on a feature in HBase which
> relies on zookeeper for coordination.
>
> A task is removed from the tasks map. An async ZK request is pending to
> remove th
Ted,
ZK does not re order async requests. They will be executed in the
same order as you call them from the same zk handle. Does that help?
thanks
mahadev
On Fri, Dec 23, 2011 at 10:43 AM, Ted Yu wrote:
> The requests come from same zhandle.
>
> Cheers
>
> On Thu, Dec 22, 2011 at 6:09 PM, Ted Y
Sure.
That helps, Mahadev.
Can I assume that async requests from different zk handles don't have
guaranteed order of execution ?
On Mon, Dec 26, 2011 at 4:23 PM, Mahadev Konar wrote:
> Ted,
> ZK does not re order async requests. They will be executed in the
> same order as you call them from th
Well, you have a constraint on the ordering from the ordering for requests
from each separate handle.
Other than that, it is hard to see how you could get a stronger ordering.
On Mon, Dec 26, 2011 at 5:53 PM, Ted Yu wrote:
> Sure.
> That helps, Mahadev.
>
> Can I assume that async requests from
On Mon, Dec 26, 2011 at 5:53 PM, Ted Yu wrote:
> Can I assume that async requests from different zk handles don't have
> guaranteed order of execution ?
Not sure what you mean. There will be _an_ order, and you're
guaranteed that all sessions will see the same order. The order will
be the order i