[Qemu-devel] Does the event idx mechanism in virtio work in the correct way?

[Bin Wu <wu.wubin@huawei.com>]

Hi,
	The event idx in virtio is an effective way to reduce the number of interrupts 
and exits of the guest. When the guest puts an request into the virtio ring, it 
doesn't exit immediately to inform the backend. Instead, the guest checks the 
"avail" event idx to determine the notification. For example, assume that the 
guest already puts five requests in the ring and the backend gets two of these 
requests, then the "avail" event idx should be two. Now the guest puts another 
request into the ring. It checks the "avail" event idx and finds the event idx 
(equals to 2) is less than the number of available requests (equals to 6), so it 
will not notify the backend to work. The backend works in a loop to get the 
request from the ring until the ring is empty or all requests have been taken.
	However, I find some code doesn't work in the way I thought:

int virtqueue_pop(VirtQueue *vq, VirtQueueElement *elem)
{
	...
	i = head = virtqueue_get_head(vq, vq->last_avail_idx++);
     	if (vdev->guest_features & (1 << VIRTIO_RING_F_EVENT_IDX)) {
         	vring_avail_event(vq, vring_avail_idx(vq));
     	}
	...
}

In the sentence "vring_avail_event(vq, vring_avail_idx(vq));", I think the 
"avail" event idx should equal to the number of requests have been 
taken(vq->last_avail_idx), not the number of all available requests 
(vring_avail_idx(vq)). Is there any special consideration or do I just 
misunderstand the event idx?

thanks

Bin Wu