On 2018年08月02日 16:41, Toshiaki Makita wrote:
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?
The problem is it does nothing if vhost_vq_avail_empty() is true and
vhost_enble_notify() is false.
+
+ 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.
Yes.
Thanks
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