Re: [RFC v3 4/5] virtio_ring: add event idx support in packed ring

[Tiwei Bie]

On Tue, May 08, 2018 at 01:40:40PM +0800, Jason Wang wrote:
> On 2018å05æ08æ 11:05, Jason Wang wrote:
> > > 
> > > Because in virtqueue_enable_cb_delayed(), we may set an
> > > event_off which is bigger than new and both of them have
> > > wrapped. And in this case, although new is smaller than
> > > event_off (i.e. the third param -- old), new shouldn't
> > > add vq->num, and actually we are expecting a very big
> > > idx diff.
> > 
> > Yes, so to calculate distance correctly between event and new, we just
> > need to compare the warp counter and return false if it doesn't match
> > without the need to try to add vq.num here.
> > 
> > Thanks
> 
> Sorry, looks like the following should work, we need add vq.num if
> used_wrap_counter does not match:
> 
> static bool vhost_vring_packed_need_event(struct vhost_virtqueue *vq,
> ÂÂÂ ÂÂÂ ÂÂÂ ÂÂÂ ÂÂÂ Â __u16 off_wrap, __u16 new,
> ÂÂÂ ÂÂÂ ÂÂÂ ÂÂÂ ÂÂÂ Â __u16 old)
> {
> ÂÂÂ bool wrap = off_wrap >> 15;
> ÂÂÂ int off = off_wrap & ~(1 << 15);
> ÂÂÂ __u16 d1, d2;
> 
> ÂÂÂ if (wrap != vq->used_wrap_counter)
> ÂÂÂ ÂÂÂ d1 = new + vq->num - off - 1;

Just to draw your attention (maybe you have already
noticed this).

In this case (i.e. wrap != vq->used_wrap_counter),
it's also possible that (off < new) is true. Because,

when virtqueue_enable_cb_delayed_packed() is used,
`off` is calculated in driver in a way like this:

	off = vq->last_used_idx + bufs;
	if (off >= vq->vring_packed.num) {
		off -= vq->vring_packed.num;
		wrap_counter ^= 1;
	}

And when `new` (in vhost) is close to vq->num. The
vq->last_used_idx + bufs (in driver) can be bigger
than vq->vring_packed.num, and:

1. `off` will wrap;
2. wrap counters won't match;
3. off < new;

And d1 (i.e. new + vq->num - off - 1) will be a value
bigger than vq->num. I'm okay with this, although it's
a bit weird.

Best regards,
Tiwei Bie

> ÂÂÂ else
> ÂÂÂ ÂÂÂ d1 = new - off - 1;
> 
> ÂÂÂ if (new > old)
> ÂÂÂ ÂÂÂ d2 = new - old;
> ÂÂÂ else
> ÂÂÂ ÂÂÂ d2 = new + vq->num - old;
> 
> ÂÂÂ return d1 < d2;
> }
> 
> Thanks
>