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Message-ID: <20180508064409.kcn6amhsxu7nkuuc@debian>
Date: Tue, 8 May 2018 14:44:09 +0800
From: Tiwei Bie <tiwei.bie@...el.com>
To: Jason Wang <jasowang@...hat.com>
Cc: "Michael S. Tsirkin" <mst@...hat.com>,
virtualization@...ts.linux-foundation.org,
linux-kernel@...r.kernel.org, netdev@...r.kernel.org,
wexu@...hat.com, jfreimann@...hat.com
Subject: Re: [RFC v3 4/5] virtio_ring: add event idx support in packed ring
On Tue, May 08, 2018 at 01:40:40PM +0800, Jason Wang wrote:
> On 2018年05月08日 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,
> __u16 off_wrap, __u16 new,
> __u16 old)
> {
> bool wrap = off_wrap >> 15;
> int off = off_wrap & ~(1 << 15);
> __u16 d1, d2;
>
> if (wrap != vq->used_wrap_counter)
> 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
> else
> d1 = new - off - 1;
>
> if (new > old)
> d2 = new - old;
> else
> d2 = new + vq->num - old;
>
> return d1 < d2;
> }
>
> Thanks
>
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