On 09/09/2016 04:30 PM, John Ferlan wrote:
>>
>> + /* Fix job completeness reporting. If cur == end mgmt
>> + * applications think job is completed. Except when both cur
>> + * and end are zero, in which case qemu hasn't started the
>> + * job yet. */
>> + if (!info->cur && !info->end) {
We get here if qemu reports 0/0 (or if qemu reports nothing, and we end
up with 0/0 because we 0-initialized the object)...
>> + if (rawInfo->ready > 0) {
>> + info->cur = info->end = 1;
if qemu reported done (on a no-op job), then we fudge to 1/1 and the
caller knows we are done...
>> + } else if (rawInfo->ready < 0) {
>> + info->end = 1;
if qemu didn't tell us it was ready, then we fudge to 0/1.
I thought the original email thread was that if rawInfo->ready == 0
(qemu explicitly told us it is NOT done) that we want to fudge to 0/1,
and then the real question is that if qemu tells us nothing at all about
rawInfo->ready, then fudging MIGHT treat a no-op job as never ending, so
it was better to leave it at 0/0 (an application getting 0/0 when
talking to new-enough libvirt then knows it is talking to older qemu).
In other words, I think this condition is slightly better as
rawInfo->ready == 0, and leave the rawInfo->ready < 0 case as 0/0.
Or am I misremembering the results of the earlier thread?
>> + }
>
> Can info->ready == 0 ? w/ info->cur = info->end = 0
>
> If so, then we're in the same mess or some other weird condition.
>
> Seems like "ready" will be set in qemu during block_job_event_ready, so
> that would say to me that as long as the structure is allocated, ready
> will be false and conceivably info->cur = info->end = 0.
>
> Wouldn't that mean the < 0 should be <= 0
<= 0 would make both explicitly not done and no answer from qemu mean
the same thing - we fudge to 0/1 to tell the caller that it is not done.
--
Eric Blake eblake redhat com +1-919-301-3266
Libvirt virtualization library http://libvirt.org
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