Re: [PATCH net V2 2/2] veth: more robust handing of race to avoid txq getting stuck

[Toshiaki Makita <toshiaki.makita1@gmail.com>]

On 2025/10/29 19:33, Jesper Dangaard Brouer wrote:
> On 28/10/2025 15.56, Toshiaki Makita wrote:
>> On 2025/10/28 5:05, Jesper Dangaard Brouer wrote:
>>> (3) Finally, the NAPI completion check in veth_poll() is updated. If NAPI is
>>> about to complete (napi_complete_done), it now also checks if the peer TXQ
>>> is stopped. If the ring is empty but the peer TXQ is stopped, NAPI will
>>> reschedule itself. This prevents a new race where the producer stops the
>>> queue just as the consumer is finishing its poll, ensuring the wakeup is not missed.
>> ...
>>
>>> @@ -986,7 +979,8 @@ static int veth_poll(struct napi_struct *napi, int budget)
>>>       if (done < budget && napi_complete_done(napi, done)) {
>>>           /* Write rx_notify_masked before reading ptr_ring */
>>>           smp_store_mb(rq->rx_notify_masked, false);
>>> -        if (unlikely(!__ptr_ring_empty(&rq->xdp_ring))) {
>>> +        if (unlikely(!__ptr_ring_empty(&rq->xdp_ring) ||
>>> +                 (peer_txq && netif_tx_queue_stopped(peer_txq)))) {
>>
>> Not sure if this is necessary.
> 
> How sure are you that this isn't necessary?
> 
>>  From commitlog, your intention seems to be making sure to wake up the queue,
>> but you wake up the queue immediately after this hunk in the same function,
>> so isn't it guaranteed without scheduling another napi?
>>
> 
> The above code catches the case, where the ptr_ring is empty and the
> tx_queue is stopped.  It feels wrong not to reach in this case, but you
> *might* be right that it isn't strictly necessary, because below code
> will also call netif_tx_wake_queue() which *should* have a SKB stored
> that will *indirectly* trigger a restart of the NAPI.

I'm a bit confused.
Wrt (3), what you want is waking up the queue, right?
Or, what you want is actually NAPI reschedule itself?

My understanding was the former (wake up the queue).
If it's correct, (3) seems not necessary because you have already woken up the queue 
in the same function.

First NAPI
  veth_poll()
    // ptr_ring_empty() and queue_stopped()
   __napi_schedule() ... schedule second NAPI
   netif_tx_wake_queue() ... wake up the queue if queue_stopped()

Second NAPI
  veth_poll()
   netif_tx_wake_queue() ... this is what you want,
                             but the queue has been woken up in the first NAPI
                             What's the point?

> I will stare some more at the code to see if I can convince myself that
> we don't have to catch this case.
> 
> Please, also provide "How sure are you that this isn't necessary?"

I could not find the case we need (3) as I explained above.

--
Toshiaki Makita