Re: [PATCH net 5/5] net/smc: put sk reference if close work was canceled

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On 20.10.23 04:41, D. Wythe wrote:


On 10/20/23 1:40 AM, Wenjia Zhang wrote:


On 19.10.23 09:33, D. Wythe wrote:


On 10/19/23 4:26 AM, Wenjia Zhang wrote:


On 17.10.23 04:06, D. Wythe wrote:


On 10/13/23 3:04 AM, Wenjia Zhang wrote:


On 11.10.23 09:33, D. Wythe wrote:
From: "D. Wythe" <alibuda@xxxxxxxxxxxxxxxxx>

Note that we always hold a reference to sock when attempting
to submit close_work.
yes
Therefore, if we have successfully
canceled close_work from pending, we MUST release that reference
to avoid potential leaks.

Isn't the corresponding reference already released inside the smc_close_passive_work()?


Hi Wenjia,

If we successfully cancel the close work from the pending state,
it means that smc_close_passive_work() has never been executed.

You can find more details here.

/**
* cancel_work_sync - cancel a work and wait for it to finish
* @work:the work to cancel
*
* Cancel @work and wait for its execution to finish. This function
* can be used even if the work re-queues itself or migrates to
* another workqueue. On return from this function, @work is
* guaranteed to be not pending or executing on any CPU.
*
* cancel_work_sync(&delayed_work->work) must not be used for
* delayed_work's. Use cancel_delayed_work_sync() instead.
*
* The caller must ensure that the workqueue on which @work was last
* queued can't be destroyed before this function returns.
*
* Return:
* %true if @work was pending, %false otherwise.
*/
boolcancel_work_sync(structwork_struct *work)
{
return__cancel_work_timer(work, false);
}

Best wishes,
D. Wythe
As I understand, queue_work() would wake up the work if the work is not already on the queue. And the sock_hold() is just prio to the queue_work(). That means, cancel_work_sync() would cancel the work either before its execution or after. If your fix refers to the former case, at this moment, I don't think the reference can be hold, thus it is unnecessary to put it.


I am quite confuse about why you think when we cancel the work before its execution,
the reference can not be hold ?


Perhaps the following diagram can describe the problem in better way :

smc_close_cancel_work
smc_cdc_msg_recv_action


sock_hold
queue_work
if (cancel_work_sync())        // successfully cancel before execution
sock_put()                        //  need to put it since we already hold a ref before   queue_work()


ha, I already thought you might ask such question:P

I think here two Problems need to be clarified:

1) Do you think the bh_lock_sock/bh_unlock_sock in the smc_cdc_msg_recv does not protect the smc_cdc_msg_recv_action() from cancel_work_sync()? Maybe that would go back to the discussion in the other patch on the behaviors of the locks.


Yes. bh_lock_sock/bh_unlock_sock can not block code execution protected by lock_sock/unlock(). That is to say, they are not exclusive.

No, the logic of the inference is very vague to me. My understand is completely different. That is what I read from the kernel code. They are not *completely* exclusive, because while the bottom half context holds the lock i.e. bh_lock_sock, the process context can not get the lock by lock_sock. (This is actually my main point of my argument for these fixes, and I didn't see any clarify from you). However, while the process context holds the lock by lock_sock, the bottom half context can still get it by bh_lock_sock, this is just like what you showed in the code in lock_sock. Once it gets the ownership, it release the spinlock.

We can use a very simple example to infer that since bh_lock_sock is type of spin-lock, if bh_lock_sock/bh_unlock_sock can block lock_sock/unlock(),
then lock_sock/unlock() can also block bh_lock_sock/bh_unlock_sock.

If this is true, when the process context already lock_sock(), the interrupt context must wait for the process to call
release_sock(). Obviously, this is very unreasonable.


2) If the queue_work returns true, as I said in the last main, the work should be (being) executed. How could the cancel_work_sync() cancel the work before execution successgully?

No, that's not true. In fact, if queue_work returns true, it simply means that we have added the task to the queue and may schedule a worker to execute it, but it does not guarantee that the task will be executed or is being executed when it returns true,
the task might still in the list and waiting some worker to execute it.

We can make a simple inference,

1. A known fact is that if no special flag (WORK_UNBOUND) is given, tasks submitted will eventually be executed on the CPU where they were submitted.

2. If the queue_work returns true, the work should be or is being executed

If all of the above are true, when we invoke queue_work in an interrupt context, does it mean that the submitted task will be executed in the interrupt context?


Best wishes,
D. Wythe

If you say the thread is not gauranteed to be waken up in then queue_work to execute the work, please explain what the kick_pool function does.

However, the spin_lock understanding is still the key problem in the cases. As I said, if it is not get clarify, we don't really need to go on to disucss this.


Fixes: 42bfba9eaa33 ("net/smc: immediate termination for SMCD link groups")
Signed-off-by: D. Wythe <alibuda@xxxxxxxxxxxxxxxxx>
---
  net/smc/smc_close.c | 3 ++-
  1 file changed, 2 insertions(+), 1 deletion(-)

diff --git a/net/smc/smc_close.c b/net/smc/smc_close.c
index 449ef45..10219f5 100644
--- a/net/smc/smc_close.c
+++ b/net/smc/smc_close.c
@@ -116,7 +116,8 @@ static void smc_close_cancel_work(struct smc_sock *smc)
      struct sock *sk = &smc->sk;
        release_sock(sk);
-    cancel_work_sync(&smc->conn.close_work);
+    if (cancel_work_sync(&smc->conn.close_work))
+        sock_put(sk);
cancel_delayed_work_sync(&smc->conn.tx_work);
      lock_sock(sk);
  }







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