Leonardo Brás <leob...@redhat.com> wrote:
> On Tue, 2022-08-02 at 08:39 +0200, Juan Quintela wrote:
>> We do the send_prepare() and the fill of the head packet without the
>> mutex held. It will help a lot for compression and later in the
>> series for zero pages.
>>
>> Notice that we can use p->pages without holding p->mutex because
>> p->pending_job == 1.
>>
>> Signed-off-by: Juan Quintela <quint...@redhat.com>
>> ---
>> migration/multifd.h | 2 ++
>> migration/multifd.c | 11 ++++++-----
>> 2 files changed, 8 insertions(+), 5 deletions(-)
>>
>> diff --git a/migration/multifd.h b/migration/multifd.h
>> index a67cefc0a2..cd389d18d2 100644
>> --- a/migration/multifd.h
>> +++ b/migration/multifd.h
>> @@ -109,7 +109,9 @@ typedef struct {
>> /* array of pages to sent.
>> * The owner of 'pages' depends of 'pending_job' value:
>> * pending_job == 0 -> migration_thread can use it.
>> + * No need for mutex lock.
>> * pending_job != 0 -> multifd_channel can use it.
>> + * No need for mutex lock.
>> */
>> MultiFDPages_t *pages;
>>
>> diff --git a/migration/multifd.c b/migration/multifd.c
>> index 09a40a9135..68fc9f8e88 100644
>> --- a/migration/multifd.c
>> +++ b/migration/multifd.c
>> @@ -663,6 +663,8 @@ static void *multifd_send_thread(void *opaque)
>> p->flags |= MULTIFD_FLAG_SYNC;
>> p->sync_needed = false;
>> }
>> + qemu_mutex_unlock(&p->mutex);
>> +
>
> If it unlocks here, we will have unprotected:
> for (int i = 0; i < p->pages->num; i++) {
> p->normal[p->normal_num] = p->pages->offset[i];
> p->normal_num++;
> }
>
> And p->pages seems to be in the mutex-protected area.
> Should it be ok?
>From the documentation:
/* array of pages to sent.
* The owner of 'pages' depends of 'pending_job' value:
* pending_job == 0 -> migration_thread can use it.
* No need for mutex lock.
* pending_job != 0 -> multifd_channel can use it.
* No need for mutex lock.
*/
MultiFDPages_t *pages;
So, it is right.
> Also, under that we have:
> if (p->normal_num) {
> ret = multifd_send_state->ops->send_prepare(p, &local_err);
> if (ret != 0) {
> qemu_mutex_unlock(&p->mutex);
> break;
> }
> }
>
> Calling mutex_unlock() here, even though the unlock already happened before,
> could cause any issue?
Good catch. Never got an error there.
Removing that bit.
> Best regards,
Thanks, Juan.
