On 23-03-21, 22:19, Jie Deng wrote:
> +static int virtio_i2c_xfer(struct i2c_adapter *adap, struct i2c_msg *msgs, int num)
> +{
> + struct virtio_i2c *vi = i2c_get_adapdata(adap);
> + struct virtqueue *vq = vi->vq;
> + struct virtio_i2c_req *reqs;
> + unsigned long time_left;
> + int ret, nr;
> +
> + reqs = kcalloc(num, sizeof(*reqs), GFP_KERNEL);
> + if (!reqs)
> + return -ENOMEM;
> +
> + mutex_lock(&vi->lock);
> +
> + ret = virtio_i2c_send_reqs(vq, reqs, msgs, num);
> + if (ret == 0)
> + goto err_unlock_free;
> +
> + nr = ret;
> + reinit_completion(&vi->completion);
> + virtqueue_kick(vq);
Coming back to this again, what is the expectation from the other side for this
? I mean there is no obvious relation between the *msgs* which we are going to
transfer (from the other side's or host's point of view). When should the host
OS call its virtqueue_kick() counterpart ?
Lemme give an example for this. Lets say that we need to transfer 3 messages
here in this routine. What we did was we prepared virtqueue for all 3 messages
together and then called virtqueue_kick().
Now if the other side (host) processes the first message and sends its reply
(with virtqueue_kick() counterpart) before processing the other two messages,
then it will end up calling virtio_i2c_msg_done() here. That will make us call
virtio_i2c_complete_reqs(), while only the first messages is processed until
now and so we will fail for the other two messages straight away.
Should we send only 1 message from i2c-virtio linux driver and then wait for
virtio_i2c_msg_done() to be called, before sending the next message to make sure
it doesn't break ?
--
viresh
