Re: [PATCH 5/9] rpc: call release_pipe only on last close

["J. Bruce Fields" <bfields@fieldses.org>]

On Mon, Nov 10, 2008 at 03:01:19PM -0500, Trond Myklebust wrote:
> On Mon, 2008-11-10 at 14:49 -0500, J. Bruce Fields wrote:
> > On Mon, Nov 10, 2008 at 02:11:37PM -0500, Trond Myklebust wrote:
> > > On Sun, 2008-11-09 at 16:04 -0500, J. Bruce Fields wrote:
> > > > I can't see any reason we particularly need to call this until the last
> > > > gssd closes the pipe.
> > > 
> > > There's a very good reason: if we call rpc_close_pipes() then it is
> > > because the kernel listener is shutting down. At that point, we want to
> > > return EPIPE for all future read() or write() attempts by gssd.
> > 
> > Sure.  This doesn't interfere with that--we're still calling
> > release_pipe() itself (which is the function that sets rpci->ops to
> > NULL), it's only the ->release_pipe() op that we're not calling.
> >
> > > > Also, this allows to guarantee that open_pipe and release_pipe are
> > > > called strictly in pairs; open_pipe on the first open, release_pipe on
> > > > the last close.  That'll make it very easy for the gss code to keep
> > > > track of which pipes gssd is using.
> > > 
> > > ...unless the gss code is no longer running.
> > 
> > I'm not following you.
> > 
> > Note the patch calls ->release_pipe() at close_pipe() time if there are
> > still opens (and then subsequent closes become no-ops).  So it does
> > guarantee that there's always a matching call.
> 
> No it doesn't. Now that rpci->ops has been set to NULL by
> rpc_close_pipes(), exactly how are you going to call
> rpci->ops->release_pipe() at a later time?

Again, I'm dealing with that case by calling release_pipe() from
rpc_close_pipes(), just as the current code does--the only change being
to do that only when there are still opens:

		last_close = rpci->nreaders != 0 || rpci->nwriters != 0;
 		rpci->nreaders = 0;
		...
 		rpci->nwriters = 0;
		if (last_close && ops->release_pipe)
 			ops->release_pipe(inode);

--b.