fuse: when are release requests queued?

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fuse: when are release requests queued?

Nikolaus Rath
Hello,

I am trying to debug a sporadic test failure in libfuse
(https://github.com/libfuse/libfuse/issues/157).


Can someone tell me at which point the fuse kernel module will send a
RELEASE request to userspace? Is it possible that this is delayed until
after the close() syscall for the last fd has returned and userspace has
submitted a different fuse request for the same fs?


Looking at fs/fuse/file.c, it looks as if fuse_release() directly calls
fuse_request_send_background() to send the request. But at that point I
can no longer follow the code. Is it possible for another request to
sneak in at this point?

Furthermore, does the VFS call fuse_release() directly while handling
the close() syscall, or does this happen asynchronously later on?


Thanks!
-Nikolaus

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Re: fuse: when are release requests queued?

David Butterfield
> Looking at fs/fuse/file.c, it looks as if fuse_release() directly calls
> fuse_request_send_background() to send the request. But at that point I
> can no longer follow the code. Is it possible for another request to
> sneak in at this point?
>
> Furthermore, does the VFS call fuse_release() directly while handling
> the close() syscall, or does this happen asynchronously later on?

Does the comment in fuse_release_common (called by fuse_release)
(Linux 4.4.0) answer this?

 267         /*
 268          * Normally this will send the RELEASE request, however if
 269          * some asynchronous READ or WRITE requests are outstanding,
 270          * the sending will be delayed.
 271          *
 272          * Make the release synchronous if this is a fuseblk mount,
 273          * synchronous RELEASE is allowed (and desirable) in this case
 274          * because the server can be trusted not to screw up.
 275          */

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Re: fuse: when are release requests queued?

Nikolaus Rath
On May 26 2017, David Butterfield <[hidden email]> wrote:

>> Looking at fs/fuse/file.c, it looks as if fuse_release() directly calls
>> fuse_request_send_background() to send the request. But at that point I
>> can no longer follow the code. Is it possible for another request to
>> sneak in at this point?
>>
>> Furthermore, does the VFS call fuse_release() directly while handling
>> the close() syscall, or does this happen asynchronously later on?
>
> Does the comment in fuse_release_common (called by fuse_release)
> (Linux 4.4.0) answer this?
>
>  267         /*
>  268          * Normally this will send the RELEASE request, however if
>  269          * some asynchronous READ or WRITE requests are outstanding,
>  270          * the sending will be delayed.
>  271          *
>  272          * Make the release synchronous if this is a fuseblk mount,
>  273          * synchronous RELEASE is allowed (and desirable) in this case
>  274          * because the server can be trusted not to screw up.
>  275          */


It does give some indication, I'd rather have someone familiar with the
actual code confirm this.

Specifically, this says that if async read()/write() requests are
pending, the RELEASE will  be delayed. But does this guarantee that
that if there are no pending requests it will not be delayed? And how
can there be a pending request if the file isn't open anymore?

Best,
-Nikolaus

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Re: fuse: when are release requests queued?

Maxim Patlasov-3
In reply to this post by Nikolaus Rath
On 05/25/2017 04:08 PM, Nikolaus Rath wrote:

> Hello,
>
> I am trying to debug a sporadic test failure in libfuse
> (https://github.com/libfuse/libfuse/issues/157).
>
>
> Can someone tell me at which point the fuse kernel module will send a
> RELEASE request to userspace?

Anytime after fuse_release(). It only puts request to background queue.
Later, the request will be transferred to pending queue. And later, the
userspace will fetch it by fuse_dev_do_read().

> Is it possible that this is delayed until
> after the close() syscall for the last fd has returned and userspace has
> submitted a different fuse request for the same fs?

I think it's possible. See how flush_bg_queue() do nothing if
fc->active_background > fc->max_background.

>
>
> Looking at fs/fuse/file.c, it looks as if fuse_release() directly calls
> fuse_request_send_background() to send the request. But at that point I
> can no longer follow the code. Is it possible for another request to
> sneak in at this point?

Yes, but not for that given fuse_file that we're closing now.

>
> Furthermore, does the VFS call fuse_release() directly while handling
> the close() syscall, or does this happen asynchronously later on?

It's called directly for well-behaved applications in well-controlled
environment, but there are some exceptions. You may be interested to
read https://sourceforge.net/p/fuse/mailman/message/32872225/


>
>
> Thanks!
> -Nikolaus
>


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Re: fuse: when are release requests queued?

Maxim Patlasov-3
In reply to this post by Nikolaus Rath
On 05/26/2017 04:11 PM, Nikolaus Rath wrote:

> On May 26 2017, David Butterfield <[hidden email]> wrote:
>>> Looking at fs/fuse/file.c, it looks as if fuse_release() directly calls
>>> fuse_request_send_background() to send the request. But at that point I
>>> can no longer follow the code. Is it possible for another request to
>>> sneak in at this point?
>>>
>>> Furthermore, does the VFS call fuse_release() directly while handling
>>> the close() syscall, or does this happen asynchronously later on?
>> Does the comment in fuse_release_common (called by fuse_release)
>> (Linux 4.4.0) answer this?
>>
>>   267         /*
>>   268          * Normally this will send the RELEASE request, however if
>>   269          * some asynchronous READ or WRITE requests are outstanding,
>>   270          * the sending will be delayed.
>>   271          *
>>   272          * Make the release synchronous if this is a fuseblk mount,
>>   273          * synchronous RELEASE is allowed (and desirable) in this case
>>   274          * because the server can be trusted not to screw up.
>>   275          */
>
> It does give some indication, I'd rather have someone familiar with the
> actual code confirm this.
>
> Specifically, this says that if async read()/write() requests are
> pending, the RELEASE will  be delayed. But does this guarantee that
> that if there are no pending requests it will not be delayed?

If nothing is pending, it will go to pending queue immediately. But this
won't guarantee that the userspace fetches it before fuse_release() returns.

> And how
> can there be a pending request if the file isn't open anymore?

I think the comment tells us about pending request in general, not
specifically for that given file.


>
> Best,
> -Nikolaus
>


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Re: fuse: when are release requests queued?

Nikolaus Rath
In reply to this post by Maxim Patlasov-3
Hi Maxim,

On May 26 2017, Maxim Patlasov <[hidden email]> wrote:

> On 05/25/2017 04:08 PM, Nikolaus Rath wrote:
>
>> Hello,
>>
>> I am trying to debug a sporadic test failure in libfuse
>> (https://github.com/libfuse/libfuse/issues/157).
>>
>>
>> Can someone tell me at which point the fuse kernel module will send a
>> RELEASE request to userspace?
>
> Anytime after fuse_release(). It only puts request to background
> queue. Later, the request will be transferred to pending queue. And
> later, the userspace will fetch it by fuse_dev_do_read().
>
>> Is it possible that this is delayed until
>> after the close() syscall for the last fd has returned and userspace has
>> submitted a different fuse request for the same fs?
>
> I think it's possible. See how flush_bg_queue() do nothing if
> fc->active_background > fc->max_background.

Thanks Maxim! Not sure what I'd do with these issues without you :-).


Is there a way to deliberate trigger this behavior for debugging? For
example, is there a kernel equivalent of sleep(1) that I could put into
fuse_release()?

>> Looking at fs/fuse/file.c, it looks as if fuse_release() directly calls
>> fuse_request_send_background() to send the request. But at that point I
>> can no longer follow the code. Is it possible for another request to
>> sneak in at this point?
>
> Yes, but not for that given fuse_file that we're closing now.

I assume that a fuse_file refers to the (formerly) opened file, right?
So e.g. a unlink() request for the same directly entry could still go
through before RELEASE has been transferred to the pending queue?


Best,
-Nikolaus

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Re: fuse: when are release requests queued?

Maxim Patlasov-3
On 05/29/2017 09:49 AM, Nikolaus Rath wrote:

> Hi Maxim,
>
> On May 26 2017, Maxim Patlasov <[hidden email]> wrote:
>> On 05/25/2017 04:08 PM, Nikolaus Rath wrote:
>>
>>> Hello,
>>>
>>> I am trying to debug a sporadic test failure in libfuse
>>> (https://github.com/libfuse/libfuse/issues/157).
>>>
>>>
>>> Can someone tell me at which point the fuse kernel module will send a
>>> RELEASE request to userspace?
>> Anytime after fuse_release(). It only puts request to background
>> queue. Later, the request will be transferred to pending queue. And
>> later, the userspace will fetch it by fuse_dev_do_read().
>>
>>> Is it possible that this is delayed until
>>> after the close() syscall for the last fd has returned and userspace has
>>> submitted a different fuse request for the same fs?
>> I think it's possible. See how flush_bg_queue() do nothing if
>> fc->active_background > fc->max_background.
> Thanks Maxim! Not sure what I'd do with these issues without you :-).
>
>
> Is there a way to deliberate trigger this behavior for debugging? For
> example, is there a kernel equivalent of sleep(1) that I could put into
> fuse_release()?

schedule_timeout_interruptible(HZ). But it's better to instrument fuse
userspace to postpone processing some i/o requests. Then you'll keep
fc->active_background > fc->max_background for a while. During that
period fuse_release may succeed with FUSE_RELEASE queued, but not passed
to the userspace. Then you cat try to sneak another request -- something
not involving fuse background queue.

>
>>> Looking at fs/fuse/file.c, it looks as if fuse_release() directly calls
>>> fuse_request_send_background() to send the request. But at that point I
>>> can no longer follow the code. Is it possible for another request to
>>> sneak in at this point?
>> Yes, but not for that given fuse_file that we're closing now.
> I assume that a fuse_file refers to the (formerly) opened file, right?
> So e.g. a unlink() request for the same directly entry could still go
> through before RELEASE has been transferred to the pending queue?

Yes. But FUSE_UNLINK works on file path, it doesn't depend on fuse_file
at all. Hence unlink request can go anytime.


>
>
> Best,
> -Nikolaus
>


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Re: fuse: when are release requests queued?

Nikolaus Rath
On May 31 2017, Maxim Patlasov <[hidden email]> wrote:

>>>> Can someone tell me at which point the fuse kernel module will send a
>>>> RELEASE request to userspace?
>>>
>>> Anytime after fuse_release(). It only puts request to background
>>> queue. Later, the request will be transferred to pending queue. And
>>> later, the userspace will fetch it by fuse_dev_do_read().
>>>
>>>> Is it possible that this is delayed until
>>>> after the close() syscall for the last fd has returned and userspace has
>>>> submitted a different fuse request for the same fs?
>>> I think it's possible. See how flush_bg_queue() do nothing if
>>> fc->active_background > fc->max_background.
>> Thanks Maxim! Not sure what I'd do with these issues without you :-).
>>
>>
>> Is there a way to deliberate trigger this behavior for debugging? For
>> example, is there a kernel equivalent of sleep(1) that I could put into
>> fuse_release()?
>
> schedule_timeout_interruptible(HZ).

Hmm. I made the following change in linux 4.10:

diff --git a/fs/fuse/file.c b/fs/fuse/file.c
index 2401c5..3568a8 100644
--- a/fs/fuse/file.c
+++ b/fs/fuse/file.c
@@ -252,6 +252,9 @@ void fuse_release_common(struct file *file, int opcode)
        if (unlikely(!ff))
                return;
 
+        // Wait a little to force race condition in userspace
+        schedule_timeout_interruptible(1);
+
        req = ff->reserved_req;
        fuse_prepare_release(ff, file->f_flags, opcode);
 

But when doing e.g. "echo test > newfile", the RELEASE request still
comes right away (judging from the libfuse debugging output).

Do I need to do something else?

> But it's better to instrument fuse
> userspace to postpone processing some i/o requests. Then you'll keep
> fc->active_background > fc->max_background for a while. During that
> period fuse_release may succeed with FUSE_RELEASE queued, but not
> passed to the userspace. Then you cat try to sneak another request --
> something not involving fuse background queue.

I don't know.. why is this better? It seems a lot more complicated. I
need to generate the extra request, add some switch to tell libfuse when
to start processing again, synchronize this with sneaking in the other
request...



Best,
-Nikolaus

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Re: fuse: when are release requests queued?

Maxim Patlasov-3
On 05/31/2017 12:19 PM, Nikolaus Rath wrote:

> On May 31 2017, Maxim Patlasov <[hidden email]> wrote:
>>>>> Can someone tell me at which point the fuse kernel module will send a
>>>>> RELEASE request to userspace?
>>>> Anytime after fuse_release(). It only puts request to background
>>>> queue. Later, the request will be transferred to pending queue. And
>>>> later, the userspace will fetch it by fuse_dev_do_read().
>>>>
>>>>> Is it possible that this is delayed until
>>>>> after the close() syscall for the last fd has returned and userspace has
>>>>> submitted a different fuse request for the same fs?
>>>> I think it's possible. See how flush_bg_queue() do nothing if
>>>> fc->active_background > fc->max_background.
>>> Thanks Maxim! Not sure what I'd do with these issues without you :-).
>>>
>>>
>>> Is there a way to deliberate trigger this behavior for debugging? For
>>> example, is there a kernel equivalent of sleep(1) that I could put into
>>> fuse_release()?
>> schedule_timeout_interruptible(HZ).
> Hmm. I made the following change in linux 4.10:
>
> diff --git a/fs/fuse/file.c b/fs/fuse/file.c
> index 2401c5..3568a8 100644
> --- a/fs/fuse/file.c
> +++ b/fs/fuse/file.c
> @@ -252,6 +252,9 @@ void fuse_release_common(struct file *file, int opcode)
>          if (unlikely(!ff))
>                  return;
>  
> +        // Wait a little to force race condition in userspace
> +        schedule_timeout_interruptible(1);
> +
>          req = ff->reserved_req;
>          fuse_prepare_release(ff, file->f_flags, opcode);
>  
>
> But when doing e.g. "echo test > newfile", the RELEASE request still
> comes right away (judging from the libfuse debugging output).
>
> Do I need to do something else?

Try HZ*10 instead of 1 as an argument of schedule_timeout_interruptible.

>
>> But it's better to instrument fuse
>> userspace to postpone processing some i/o requests. Then you'll keep
>> fc->active_background > fc->max_background for a while. During that
>> period fuse_release may succeed with FUSE_RELEASE queued, but not
>> passed to the userspace. Then you cat try to sneak another request --
>> something not involving fuse background queue.
> I don't know.. why is this better? It seems a lot more complicated. I
> need to generate the extra request, add some switch to tell libfuse when
> to start processing again, synchronize this with sneaking in the other
> request...

I thought it's better because it would trigger delayed processing of
FUSE_RELEASE: last __fput() succeeded, but fuse userspace will see
FUSE_RELEASE only later. Adding sleep to fuse_release_common would only
extend processing time of last __fput(), is that what you need?

>
>
>
> Best,
> -Nikolaus
>


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Re: fuse: when are release requests queued?

Nikolaus Rath
On May 31 2017, Maxim Patlasov <[hidden email]> wrote:

> On 05/31/2017 12:19 PM, Nikolaus Rath wrote:
>
>> On May 31 2017, Maxim Patlasov <[hidden email]> wrote:
>>>>>> Can someone tell me at which point the fuse kernel module will send a
>>>>>> RELEASE request to userspace?
>>>>> Anytime after fuse_release(). It only puts request to background
>>>>> queue. Later, the request will be transferred to pending queue. And
>>>>> later, the userspace will fetch it by fuse_dev_do_read().
>>>>>
>>>>>> Is it possible that this is delayed until
>>>>>> after the close() syscall for the last fd has returned and userspace has
>>>>>> submitted a different fuse request for the same fs?
>>>>> I think it's possible. See how flush_bg_queue() do nothing if
>>>>> fc->active_background > fc->max_background.
>>>> Thanks Maxim! Not sure what I'd do with these issues without you :-).
>>>>
>>>>
>>>> Is there a way to deliberate trigger this behavior for debugging? For
>>>> example, is there a kernel equivalent of sleep(1) that I could put into
>>>> fuse_release()?
>>> schedule_timeout_interruptible(HZ).
>> Hmm. I made the following change in linux 4.10:
>>
>> diff --git a/fs/fuse/file.c b/fs/fuse/file.c
>> index 2401c5..3568a8 100644
>> --- a/fs/fuse/file.c
>> +++ b/fs/fuse/file.c
>> @@ -252,6 +252,9 @@ void fuse_release_common(struct file *file, int opcode)
>>          if (unlikely(!ff))
>>                  return;
>>   +        // Wait a little to force race condition in userspace
>> +        schedule_timeout_interruptible(1);
>> +
>>          req = ff->reserved_req;
>>          fuse_prepare_release(ff, file->f_flags, opcode);
>>  
>>
>> But when doing e.g. "echo test > newfile", the RELEASE request still
>> comes right away (judging from the libfuse debugging output).
>>
>> Do I need to do something else?
>
> Try HZ*10 instead of 1 as an argument of
> schedule_timeout_interruptible.

Ok, now the RELEASE comes a lot later. But now userspace is also
blocking until RELEASE comes in.

>>> But it's better to instrument fuse
>>> userspace to postpone processing some i/o requests. Then you'll keep
>>> fc->active_background > fc->max_background for a while. During that
>>> period fuse_release may succeed with FUSE_RELEASE queued, but not
>>> passed to the userspace. Then you cat try to sneak another request --
>>> something not involving fuse background queue.
>>
>> I don't know.. why is this better? It seems a lot more complicated. I
>> need to generate the extra request, add some switch to tell libfuse when
>> to start processing again, synchronize this with sneaking in the other
>> request...
>
> I thought it's better because it would trigger delayed processing of
> FUSE_RELEASE: last __fput() succeeded, but fuse userspace will see
> FUSE_RELEASE only later. Adding sleep to fuse_release_common would
> only extend processing time of last __fput(), is that what you need?

I do not fully understand the difference you describe. What I would like
to construct is the following scenario:

1. Userspace calls close()
2. Userspace close() returns
3. Userspace calls unlink()
4. Userspace unlink() returns
5. libfuse reads UNLINK request from kernel pipe
6. libfuse reads RELEASE request from kernel pipe

What would be the simplest way to do that?

Thanks!
-Nikolaus

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Re: fuse: when are release requests queued?

Michael Theall-2
Won't unlink(2) block until the fuse server has responded? I'm pretty sure the close(2) should come back after the fuse server responds to FLUSH. It sounds like with your RELEASE delay in the kernel, you should get your steps as described buy step 4 and 5 must be swapped.

Regards,
Michael Theall

On Wed, May 31, 2017 at 2:41 PM Nikolaus Rath <[hidden email]> wrote:
On May 31 2017, Maxim Patlasov <[hidden email]> wrote:
> On 05/31/2017 12:19 PM, Nikolaus Rath wrote:
>
>> On May 31 2017, Maxim Patlasov <[hidden email]> wrote:
>>>>>> Can someone tell me at which point the fuse kernel module will send a
>>>>>> RELEASE request to userspace?
>>>>> Anytime after fuse_release(). It only puts request to background
>>>>> queue. Later, the request will be transferred to pending queue. And
>>>>> later, the userspace will fetch it by fuse_dev_do_read().
>>>>>
>>>>>> Is it possible that this is delayed until
>>>>>> after the close() syscall for the last fd has returned and userspace has
>>>>>> submitted a different fuse request for the same fs?
>>>>> I think it's possible. See how flush_bg_queue() do nothing if
>>>>> fc->active_background > fc->max_background.
>>>> Thanks Maxim! Not sure what I'd do with these issues without you :-).
>>>>
>>>>
>>>> Is there a way to deliberate trigger this behavior for debugging? For
>>>> example, is there a kernel equivalent of sleep(1) that I could put into
>>>> fuse_release()?
>>> schedule_timeout_interruptible(HZ).
>> Hmm. I made the following change in linux 4.10:
>>
>> diff --git a/fs/fuse/file.c b/fs/fuse/file.c
>> index 2401c5..3568a8 100644
>> --- a/fs/fuse/file.c
>> +++ b/fs/fuse/file.c
>> @@ -252,6 +252,9 @@ void fuse_release_common(struct file *file, int opcode)
>>          if (unlikely(!ff))
>>                  return;
>>   +        // Wait a little to force race condition in userspace
>> +        schedule_timeout_interruptible(1);
>> +
>>          req = ff->reserved_req;
>>          fuse_prepare_release(ff, file->f_flags, opcode);
>>
>>
>> But when doing e.g. "echo test > newfile", the RELEASE request still
>> comes right away (judging from the libfuse debugging output).
>>
>> Do I need to do something else?
>
> Try HZ*10 instead of 1 as an argument of
> schedule_timeout_interruptible.

Ok, now the RELEASE comes a lot later. But now userspace is also
blocking until RELEASE comes in.

>>> But it's better to instrument fuse
>>> userspace to postpone processing some i/o requests. Then you'll keep
>>> fc->active_background > fc->max_background for a while. During that
>>> period fuse_release may succeed with FUSE_RELEASE queued, but not
>>> passed to the userspace. Then you cat try to sneak another request --
>>> something not involving fuse background queue.
>>
>> I don't know.. why is this better? It seems a lot more complicated. I
>> need to generate the extra request, add some switch to tell libfuse when
>> to start processing again, synchronize this with sneaking in the other
>> request...
>
> I thought it's better because it would trigger delayed processing of
> FUSE_RELEASE: last __fput() succeeded, but fuse userspace will see
> FUSE_RELEASE only later. Adding sleep to fuse_release_common would
> only extend processing time of last __fput(), is that what you need?

I do not fully understand the difference you describe. What I would like
to construct is the following scenario:

1. Userspace calls close()
2. Userspace close() returns
3. Userspace calls unlink()
4. Userspace unlink() returns
5. libfuse reads UNLINK request from kernel pipe
6. libfuse reads RELEASE request from kernel pipe

What would be the simplest way to do that?

Thanks!
-Nikolaus

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Re: fuse: when are release requests queued?

Maxim Patlasov-3
In reply to this post by Nikolaus Rath
On 05/31/2017 12:41 PM, Nikolaus Rath wrote:

> On May 31 2017, Maxim Patlasov <[hidden email]> wrote:
>> On 05/31/2017 12:19 PM, Nikolaus Rath wrote:
>>
>>> On May 31 2017, Maxim Patlasov <[hidden email]> wrote:
>>>>>>> Can someone tell me at which point the fuse kernel module will send a
>>>>>>> RELEASE request to userspace?
>>>>>> Anytime after fuse_release(). It only puts request to background
>>>>>> queue. Later, the request will be transferred to pending queue. And
>>>>>> later, the userspace will fetch it by fuse_dev_do_read().
>>>>>>
>>>>>>> Is it possible that this is delayed until
>>>>>>> after the close() syscall for the last fd has returned and userspace has
>>>>>>> submitted a different fuse request for the same fs?
>>>>>> I think it's possible. See how flush_bg_queue() do nothing if
>>>>>> fc->active_background > fc->max_background.
>>>>> Thanks Maxim! Not sure what I'd do with these issues without you :-).
>>>>>
>>>>>
>>>>> Is there a way to deliberate trigger this behavior for debugging? For
>>>>> example, is there a kernel equivalent of sleep(1) that I could put into
>>>>> fuse_release()?
>>>> schedule_timeout_interruptible(HZ).
>>> Hmm. I made the following change in linux 4.10:
>>>
>>> diff --git a/fs/fuse/file.c b/fs/fuse/file.c
>>> index 2401c5..3568a8 100644
>>> --- a/fs/fuse/file.c
>>> +++ b/fs/fuse/file.c
>>> @@ -252,6 +252,9 @@ void fuse_release_common(struct file *file, int opcode)
>>>           if (unlikely(!ff))
>>>                   return;
>>>    +        // Wait a little to force race condition in userspace
>>> +        schedule_timeout_interruptible(1);
>>> +
>>>           req = ff->reserved_req;
>>>           fuse_prepare_release(ff, file->f_flags, opcode);
>>>    
>>>
>>> But when doing e.g. "echo test > newfile", the RELEASE request still
>>> comes right away (judging from the libfuse debugging output).
>>>
>>> Do I need to do something else?
>> Try HZ*10 instead of 1 as an argument of
>> schedule_timeout_interruptible.
> Ok, now the RELEASE comes a lot later. But now userspace is also
> blocking until RELEASE comes in.

That's exactly why I thought that adding sleep there wouldn't be very
useful.

>
>>>> But it's better to instrument fuse
>>>> userspace to postpone processing some i/o requests. Then you'll keep
>>>> fc->active_background > fc->max_background for a while. During that
>>>> period fuse_release may succeed with FUSE_RELEASE queued, but not
>>>> passed to the userspace. Then you cat try to sneak another request --
>>>> something not involving fuse background queue.
>>> I don't know.. why is this better? It seems a lot more complicated. I
>>> need to generate the extra request, add some switch to tell libfuse when
>>> to start processing again, synchronize this with sneaking in the other
>>> request...
>> I thought it's better because it would trigger delayed processing of
>> FUSE_RELEASE: last __fput() succeeded, but fuse userspace will see
>> FUSE_RELEASE only later. Adding sleep to fuse_release_common would
>> only extend processing time of last __fput(), is that what you need?
> I do not fully understand the difference you describe. What I would like
> to construct is the following scenario:
>
> 1. Userspace calls close()
> 2. Userspace close() returns
> 3. Userspace calls unlink()
> 4. Userspace unlink() returns
> 5. libfuse reads UNLINK request from kernel pipe
> 6. libfuse reads RELEASE request from kernel pipe
>
> What would be the simplest way to do that?

I would try to keep fc->active_background elevated somehow. For example
you add sleep(1) for every incoming write request to libfuse and
serialize processing them. Then you generate enough writes to achieve
fc->max_background. If you call close() now, and if it really ends up in
last __fput(), corresponding FUSE_RELEASE will sit in background queue
for long while (as many seconds as # elements in the queue). But close()
from your 2. will return much earlier because it doesn't wait for
completion of FUSE_RELEASE. Hence unlink() might succeed.

>
> Thanks!
> -Nikolaus
>


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Re: fuse: when are release requests queued?

Nikolaus Rath
On May 31 2017, Maxim Patlasov <[hidden email]> wrote:

>> I do not fully understand the difference you describe. What I would like
>> to construct is the following scenario:
>>
>> 1. Userspace calls close()
>> 2. Userspace close() returns
>> 3. Userspace calls unlink()
>> 4. Userspace unlink() returns
>> 5. libfuse reads UNLINK request from kernel pipe
>> 6. libfuse reads RELEASE request from kernel pipe
>>
>> What would be the simplest way to do that?
>
> I would try to keep fc->active_background elevated somehow. For
> example you add sleep(1) for every incoming write request to libfuse
> and serialize processing them. Then you generate enough writes to
> achieve fc->max_background. If you call close() now, and if it really
> ends up in last __fput(), corresponding FUSE_RELEASE will sit in
> background queue for long while (as many seconds as # elements in the
> queue). But close() from your 2. will return much earlier because it
> doesn't wait for completion of FUSE_RELEASE. Hence unlink() might
> succeed.

Ah, got it now, thanks!

Wouldn't be a simpler solution be to just patch the kernel module to
*always* put FUSE_RELEASE requests into the background queue, so that I
don't have to manually keep fc->active_background elevated?

I just can't seem to find the code that does this check... I would
expect it in fuse_file_put(), but the condition in there does not seem to
look at the number of background requests at all.


Best,
-Nikolaus

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Re: fuse: when are release requests queued?

Nikolaus Rath
In reply to this post by Michael Theall-2
On May 31 2017, Michael Theall <[hidden email]> wrote:

>> I do not fully understand the difference you describe. What I would like
>> to construct is the following scenario:
>>
>> 1. Userspace calls close()
>> 2. Userspace close() returns
>> 3. Userspace calls unlink()
>> 4. Userspace unlink() returns
>> 5. libfuse reads UNLINK request from kernel pipe
>> 6. libfuse reads RELEASE request from kernel pipe
>>
>> What would be the simplest way to do that?
>>
> Won't unlink(2) block until the fuse server has responded?

Yes, you are right. It should be:

 1. Userspace calls close()
 2. Userspace close() returns
 3. Userspace calls unlink()
 4. libfuse reads UNLINK request from kernel pipe
 5. Userspace unlink() returns
 6. libfuse reads RELEASE request from kernel pipe

> I'm pretty sure
> the close(2) should come back after the fuse server responds to FLUSH. It
> sounds like with your RELEASE delay in the kernel, you should get your
> steps as described buy step 4 and 5 must be swapped.

No, the delay comes in between (1) and (2).

Best,
-Nikolaus

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Re: fuse: when are release requests queued?

Maxim Patlasov-3
In reply to this post by Nikolaus Rath
On 05/31/2017 01:31 PM, Nikolaus Rath wrote:

> On May 31 2017, Maxim Patlasov <[hidden email]> wrote:
>>> I do not fully understand the difference you describe. What I would like
>>> to construct is the following scenario:
>>>
>>> 1. Userspace calls close()
>>> 2. Userspace close() returns
>>> 3. Userspace calls unlink()
>>> 4. Userspace unlink() returns
>>> 5. libfuse reads UNLINK request from kernel pipe
>>> 6. libfuse reads RELEASE request from kernel pipe
>>>
>>> What would be the simplest way to do that?
>> I would try to keep fc->active_background elevated somehow. For
>> example you add sleep(1) for every incoming write request to libfuse
>> and serialize processing them. Then you generate enough writes to
>> achieve fc->max_background. If you call close() now, and if it really
>> ends up in last __fput(), corresponding FUSE_RELEASE will sit in
>> background queue for long while (as many seconds as # elements in the
>> queue). But close() from your 2. will return much earlier because it
>> doesn't wait for completion of FUSE_RELEASE. Hence unlink() might
>> succeed.
> Ah, got it now, thanks!
>
> Wouldn't be a simpler solution be to just patch the kernel module to
> *always* put FUSE_RELEASE requests into the background queue, so that I
> don't have to manually keep fc->active_background elevated?
>
> I just can't seem to find the code that does this check... I would
> expect it in fuse_file_put(), but the condition in there does not seem to
> look at the number of background requests at all.

The decision is made at mount stage: it's either fuseblk mount or not.
If it's not fuseblk mount, the kernel always put FUSE_RELEASE to
background queue. And vice versa.

Keeping active_background elevated may help us to win the race: you want
unlink is queued and processed before the userspace reads FUSE_RELEASE
from kernel.


>
>
> Best,
> -Nikolaus
>


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