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Date: Fri, 8 Sep 2017 12:43:14 +0100
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From: Martin Simmons <martin@lispworks.com>
To: freebsd-fs@freebsd.org
In-reply-to: <87ingugw2v.fsf@thinkpad.rath.org> (message from Nikolaus Rath on
 Thu, 07 Sep 2017 21:14:32 +0200)
Subject: Re: umount() taking minutes for FUSE filesystems
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>>>>> On Thu, 07 Sep 2017 21:14:32 +0200, Nikolaus Rath said:
> 
> On Sep 05 2017, Martin Simmons <martin@lispworks.com> wrote:
> >> Probably the crucial difference is that the test that takes long exits
> >> its main loop on its own and then informs the FUSE kernel module about
> >> that, while the other tests terminate the main loop because the kernel
> >> module tells them to do so.
> >
> > What does "informs the FUSE kernel module about that" do to inform it?
> 
> It calls unmount.
> https://github.com/libfuse/libfuse/blob/master/lib/mount_bsd.c#L127

It seems to me that the following occurs:

1. The user program exits the main loop.
2. The user program sends unmount to the kernel.
3. The kernel sends FUSE_DESTROY to the user program.
4. The kernel waits for the user program to respond to FUSE_DESTROY.
5. The user program does not respond because it has exited the main loop.
6. The kernel wait times out after 60 seconds.

Perhaps the solution is to close the fd before calling unmount(), like the
Linux code in https://github.com/libfuse/libfuse/blob/master/lib/mount.c#L275
does?

That will prevent the kernel from sending FUSE_DESTROY, but what is the
purpose of FUSE_DESTROY if the user program can't respond to it?

__Martin