Messages in this thread | | | Date | Tue, 30 Sep 2014 21:19:33 +0200 | From | Miklos Szeredi <> | Subject | Re: [PATCH 0/5] fuse: handle release synchronously (v4) |
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On Tue, Sep 30, 2014 at 12:04:08PM -0700, Linus Torvalds wrote: > On Mon, Sep 29, 2014 at 10:33 PM, Anand Avati <avati@gluster.org> wrote: > > > > In general that sounds reasonable. The problem (as described in the original > > thread, at http://sourceforge.net/p/fuse/mailman/message/29889055/) > > From a quick look at that thread, the solution is clear: you *must* > flush your write buffers in the "flush" function. > > The fact that you must flush write buffers multiple times if people > have done "dup()" is a complete non-issue. Just flush them each time. > There is no "how do I differentiate the first flush and > the second flush?" The answer is that you don't, and that you MUST > NOT. You need to flush on both (or more). Trying to distinguish first > vs second is broken, and would be wrogn *anyway*. There is no possible > situation where it could validaly matter, and you simply cannot tell. > > Don't do any data structure cleanups, that's for "release()". But yes, > you do have to flush write buffers at flush time (and return IO errors > if they happen). That's very much the point of flush.
What about flock(2), FL_SETLEASE, etc semantics (which are the sane ones, compared to the POSIX locks shit which mandates release of lock on each close(2) instead of "when all [duplicate] descriptors have been closed")?
You have to do that from ->release(), there's no question about that. And while I haven't looked at the wording of the standards, doing that synchronously with the last close is a pretty decent thing to expect.
Thanks, Miklos
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