On 04/03/2012 04:53 PM, Anand Avati wrote:
On Tue, Apr 3, 2012 at 6:20 AM, Xavier
Hernandez <xhernandez@xxxxxxxxxx>
wrote:
Hello developers,
I'm currently trying to implement a new method for managing
inode and entry modifications that will be faster (I hope)
than the current method for the most common cases. To do so I
need to know exactly how the locking mechanism works. I have
been browsing the source code and doing some tests, and I
would like to be sure that I have understood it correctly
before continuing.
All information is based on latest qa releases from 3.3
branch.
My understanding is this:
- There are three locking fops: lk, inodelk and entrylk.
- Client application locks created using fcntl() are received
by the translators as lk requests.
- All other functionalities of lk fop are not currently used
by any translator (I mean F_RESLK_LCK, F_RESLK_LCKW,
F_RESLK_UNLCK and F_GETLK_FD).
- inodelk and entrlylk are only used by AFR to lock inodes or
directory entries before modification.
- Translators don't generate lk requests internally.
- Client application requests cannot directly generate an
inodelk or entrylk requests.
Correct so far.
- inodelk and entrylk locks are always mandatory.
inodelk and entrylk are always advisory. Never mandatory
(at least so far)
Sorry, I was thinking one thing and said another. Never mind, it's
as I expected: no request will be blocked by any of these locks
(except themselves)
- lk locks may be mandatory or advisory.
mandatory mode is not tested for a very long time.
Ok, I won't waste much time on mandatory locks.
- lk and inodelk are independent from each other, meaning that
a lock using lk will not be visible to inodelk and will not
block it. inodelk won't block lk requests neither.
- User requests can only be blocked by lk created locks
Correct, and only if mandatory locks are enabled (which
isn't tested right now)
Ok.
(if a write
request from user is allowed to pass without using inodelk, it
won't be blocked by a previous inodelk).
inodelks are never mandatory, so the question does not
apply
Ok.
Thank you very much
Xavi
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