So I think the notation in the code is misunderstanding?
Or the code to get list<pg_log_entry_t> divergent in merge_log is wrong?
// move aside divergent items
list<pg_log_entry_t> divergent;
while (!log.empty()) {
pg_log_entry_t &oe = *log.log.rbegin();
/*
* look at eversion.version here. we want to avoid a situation like:
* our log: 100'10 (0'0) m 10000004d3a.00000000/head by
client4225.1:18529
* new log: 122'10 (0'0) m 10000004d3a.00000000/head by
client4225.1:18529
* lower_bound = 100'9
* i.e, same request, different version. If the eversion.version is > the
* lower_bound, we it is divergent.
*/
if (oe.version.version <= lower_bound.version)
break;
dout(10) << "merge_log divergent " << oe << dendl;
divergent.push_front(oe);
log.log.pop_back();
}
As the logic above, divergent will contain all log_entry whose
version is the same in olog and log?
Regards
Ning Yao
2016-09-02 23:50 GMT+08:00 Samuel Just <sjust@xxxxxxxxxx>:
> On Thu, Sep 1, 2016 at 7:44 PM, Ning Yao <zay11022@xxxxxxxxx> wrote:
>> Hi, all
>>
>> we got the same issue http://tracker.ceph.com/issues/16279
>>
>> I think it is possible objiter->second->version == last_divergent_update?
>>
>> If olog is [200'5 ~ 200'10, 202'11~ 202'20] with object A: [200'5, 200'8]
>> log is [200'3 ~ 200'11], object A : [200'5, 200'8]
>>
>> Then after merge log os [200'3 ~ 200'10, 202'11~202'20] with divergent entry:
>> object A: 200'5 and 200'8
>
> I think I'm misunderstanding your notation, but if I'm reading this
> right, the entries at 200'5 and 200'8 would *not* be divergent -- the
> only divergent entry would be 200'11.
>
>>
>> first_divergent_update will be 200'5 and last_divergent_update will be
>> 200'8 for object A. If can find it in current log, and the possible
>> version the object A is equal or greater than 200'8. So I am not sure
>> it must be greater than last_divergent_update but may be equal?
>>
>> Am I missing something else?
>>
>> Regards
>> Ning Yao
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