On 2018/08/02 17:18, Jason Wang wrote:
> On 2018年08月01日 17:52, Tonghao Zhang wrote:
>>> +static void vhost_net_busy_poll_check(struct vhost_net *net,
>>> + struct vhost_virtqueue *rvq,
>>> + struct vhost_virtqueue *tvq,
>>> + bool rx)
>>> +{
>>> + struct socket *sock = rvq->private_data;
>>> +
>>> + if (rx)
>>> + vhost_net_busy_poll_try_queue(net, tvq);
>>> + else if (sock && sk_has_rx_data(sock->sk))
>>> + vhost_net_busy_poll_try_queue(net, rvq);
>>> + else {
>>> + /* On tx here, sock has no rx data, so we
>>> + * will wait for sock wakeup for rx, and
>>> + * vhost_enable_notify() is not needed. */
>>
>> A possible case is we do have rx data but guest does not refill the rx
>> queue. In this case we may lose notifications from guest.
> Yes, should consider this case. thanks.
I'm a bit confused. Isn't this covered by the previous
"else if (sock && sk_has_rx_data(...))" block?
>>>> +
>>>> + cpu_relax();
>>>> + }
>>>> +
>>>> + preempt_enable();
>>>> +
>>>> + if (!rx)
>>>> + vhost_net_enable_vq(net, vq);
>>> No need to enable rx virtqueue, if we are sure handle_rx() will be
>>> called soon.
>> If we disable rx virtqueue in handle_tx and don't send packets from
>> guest anymore(handle_tx is not called), so we can wake up for sock rx.
>> so the network is broken.
>
> Not sure I understand here. I mean is we schedule work for handle_rx(),
> there's no need to enable it since handle_rx() will do this for us.
Looks like in the last "else" block in vhost_net_busy_poll_check() we
need to enable vq since in that case we have no rx data and handle_rx()
is not scheduled.
--
Toshiaki Makita
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