On Tue, 14 Apr 2015 17:13:52 +0200
Greg Kurz <gkurz@xxxxxxxxxxxxxxxxxx> wrote:
> On Tue, 14 Apr 2015 16:20:23 +0200
> Cornelia Huck <cornelia.huck@xxxxxxxxxx> wrote:
>
> > On Fri, 10 Apr 2015 12:19:16 +0200
> > Greg Kurz <gkurz@xxxxxxxxxxxxxxxxxx> wrote:
> >
> > > +#ifdef CONFIG_VHOST_SET_ENDIAN_LEGACY
> > > +static void vhost_init_is_le(struct vhost_virtqueue *vq)
> > > +{
> > > + vq->is_le = vhost_has_feature(vq, VIRTIO_F_VERSION_1) || !vq->user_be;
> >
> > So if the endianness is not explicitly set to BE, it will be forced to
> > LE (instead of native endian)? Won't that break userspace that does not
> > yet use the new interface when CONFIG_VHOST_SET_ENDIAN_LEGACY is set?
> >
>
> If userspace doesn't use the new interface, then vq->user_be will retain its
> default value that was set in vhost_vq_reset(), i.e. :
>
> vq->user_be = !virtio_legacy_is_little_endian();
>
> This means default is native endian.
>
> What about adding this comment ?
>
> static void vhost_init_is_le(struct vhost_virtqueue *vq)
> {
> /* Note for legacy virtio: user_be is initialized in vhost_vq_reset()
> * according to the host endianness. If userspace does not set an
> * explicit endianness, the default behavior is native endian, as
> * expected by legacy virtio.
> */
Good idea, as this is easy to miss.
> vq->is_le = vhost_has_feature(vq, VIRTIO_F_VERSION_1) || !vq->user_be;
> }
>
> > > +}
> > > +#else
> > > +static void vhost_init_is_le(struct vhost_virtqueue *vq)
> > > +{
> > > + if (vhost_has_feature(vq, VIRTIO_F_VERSION_1))
> > > + vq->is_le = true;
> > > +}
> > > +#endif
> > > +
> > > int vhost_init_used(struct vhost_virtqueue *vq)
> > > {
> > > __virtio16 last_used_idx;
> > > int r;
> > > - if (!vq->private_data)
> > > + if (!vq->private_data) {
> > > + vq->is_le = virtio_legacy_is_little_endian();
> > > return 0;
> > > + }
> > > +
> > > + vhost_init_is_le(vq);
> > >
> > > r = vhost_update_used_flags(vq);
> > > if (r)
>
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