umount() taking minutes for FUSE filesystems

[Nikolaus Rath]

On Sep 08 2017, Martin Simmons <martin at lispworks.com> wrote:
>>>>>> On Thu, 07 Sep 2017 21:14:32 +0200, Nikolaus Rath said:
>> 
>> On Sep 05 2017, Martin Simmons <martin at 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 sounds reasonable, will give it a shot.

> 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?

In most cases, umount() is not called by the filesystem but by the
user. In that case, the FUSE_DESTROY request gives the filesystem a
chance to clean-up and exit the main loop.

That said, the filesystem could also detect the unmount by the kernel
closing the fd. So I'm not sure what is gained by the extra request
either... Theoretically, at least under Linux the destroy handler could
perform some notify_* operations, but I don't see how that would be
useful when the filesystem will be unmounted anyway.


Best,
-Nikolaus

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