question on sync()

question on sync()

[Surbhi Palande]

Hie,

While looking at fs/sync.c :: sync(), I found out that:

1) sync() first calls wakeup_flusher_threads() which ultimately calls
bdi_alloc_queue_work() with WB_SYNC_NONE for every bdi in the bdi_list.

2) sync() immediate then calls sync_filesystems(0) which does the same
thing - it calls writeback_inodes_sb() which finally calls
bdi_alloc_queue_work() with WB_SYNC_NONE for every bdi corresponding to
every filesystem.

If it is possible, then I wanted to know why there are effectively two
calls which ultimately call bdi_alloc_queue_work() with WB_SYNC_NONE
twice?

Thanks!

Warm Regards,
Surbhi.

Re: question on sync()

[Jan Kara]

  Hi,

> While looking at fs/sync.c :: sync(), I found out that:
> 
> 1) sync() first calls wakeup_flusher_threads() which ultimately calls
> bdi_alloc_queue_work() with WB_SYNC_NONE for every bdi in the bdi_list.
> 
> 2) sync() immediate then calls sync_filesystems(0) which does the same
> thing - it calls writeback_inodes_sb() which finally calls
> bdi_alloc_queue_work() with WB_SYNC_NONE for every bdi corresponding to
> every filesystem.
> 
> If it is possible, then I wanted to know why there are effectively two
> calls which ultimately call bdi_alloc_queue_work() with WB_SYNC_NONE
> twice?
  It's mostly a leftover from times when pdflush did background writing and
sync_filesystem() was submitting IO on it's own (i.e., not leaving the work on
flusher threads). Currently, you are right that wakeup_flusher_threads() is
superfluous. But the whole sync code would need a cleanup to properly reflect
the fact that know *all* the writes are done in background flusher threads.
For example __sync_filesystem() in wait=0 case doesn't make much sense
currently.

								Honza
-- 
Jan Kara <jack@suse.cz>
SuSE CR Labs