>
> On Tue, 2018-06-12 at 13:30 +0200, Christophe Fergeau wrote:
> > Hey,
> >
> > On Wed, May 09, 2018 at 01:20:19PM +0200, Lukáš Hrázký wrote:
> > > On Wed, 2018-05-09 at 05:18 -0400, Frediano Ziglio wrote:
> > > > The agent send a single message, but reads/writes to the device are
> > > > not atomic. Note that the current protocol introduce additional
> > > > delays as the frames cannot be partially decoded but must wait for the
> > > > full message (maybe the client can change its read code to handle this,
> > > > at the moment it does nothing about, on the server is less of a problem
> > > > as the message is build quickly from device to memory so not much delay
> > > > is added).
> > >
> > > This is the part I don't understand... AFAICS, you read the whole
> > > message in red-stream-device.c:handle_msg_data(). That should be the
> > > whole frame? Then you send the whole frame with
> > > stream_channel_send_data(). So it should never be partial?
> >
> > I would guess the data you get in handle_msg_data() got fragmented
> > somewhere on the agent -> kernel -> virtio-serial -> qemu -> spice way.
> > So one mjpeg frame will correspond to multiple calls to
> > n = sif->read(sin, buf, MIN(sizeof(buf), dev->hdr.size));
> > Then with this patch, stream_channel_send_data() will coalesce all these
> > reads in a single StreamDataItem and send it when its size corresponds
> > to dev->hdr.size.
>
> Yah, figured that out when I revisited this few days ago after a
> discussion with Uri. He mentioned he's got another patch for it, though
> I'm not sure where it is (I may have missed it).
>
> > > > > > void
> > > > > > @@ -563,11 +609,25 @@ stream_channel_send_data(StreamChannel
> > > > > > *channel,
> > > > > > const void *data, size_t size,
> > > > > >
> > > > > > RedChannel *red_channel = RED_CHANNEL(channel);
> > > > > >
> > > > > > - StreamDataItem *item = stream_data_item_new(channel, size,
> > > > > > mm_time);
> > > > > > - stream_channel_update_queue_stat(channel, 1, size);
> > > > > > - // TODO try to optimize avoiding the copy
> > > > > > - memcpy(item->data.data, data, size);
> > > > > > - red_channel_pipes_add(red_channel, &item->base);
> > > > > > + while (size) {
> > > > > > + if (channel->data_item == NULL) {
> > > > > > + stream_channel_init_data_item(channel, size, mm_time);
> > > > > > + }
> > > > > > +
> > > > > > + StreamDataItem *item = channel->data_item;
> > > > > > +
> > > > > > + size_t copy_size = item->data.data_size -
> > > > > > channel->data_item_pos;
> > > > > > + copy_size = MIN(copy_size, size);
> > > > > > + // TODO try to optimize avoiding the copy
> > > > > > + memcpy(item->data.data + channel->data_item_pos, data,
> > > > > > copy_size);
> > > > > > + size -= copy_size;
> > > > > > + channel->data_item_pos += copy_size;
> > > > > > + if (channel->data_item_pos == item->data.data_size) {
> > > > > > + channel->data_item = NULL;
> > > > > > + stream_channel_update_queue_stat(channel, 1,
> > > > > > item->data.data_size);
> > > > > > + red_channel_pipes_add(red_channel, &item->base);
> > > > > > + }
> > > > > > + }
> > >
> > > What does the while (size) loop do here? It will do more than one
> > > iteration only if copy_size < size, which means there is not enough
> > > space in the item buffer and in that case it seems to me it will loop
> > > forever? Am I missing something?
> >
> > I don't think it will loop forever (channel->data_item will be set to
> > NULL, and then a new one will be created on the next iteration, and this
> > one will be sent, and at that point, size will be 0).
>
> Same as above... figured it out, not sure what I was thinking at the
> time :)
>
> > However, I'm also
> > slightly confused as for the intent of that loop. Maybe to handle the
> > case when we receive more data than we expect?
>
> Yeah, come look at it once again (unless I'm wrong again) seems you're
> right :) I don't think we can actually receive more data for the
> message than what was the size in the header? Because the size defines
> the end of the message? So:
>
> 1. The loop should never do more than one iteration.
>
Does in case the guest wants to send very large message, in this case
is limited to 32Mb.
> 2. If it actually does, I think it will fragment the message once again
> (sending two messages for a single frame), so the bug with the client
> persists.
>
Simply not expected to be a regression in this case.
We need a single message for frame only with Mjpeg.
> I was also thinking the ideal way of handling this would be to "pipe
> through" the data, i.e. send a header with the full data message size
> to the streaming channel once a data message header arrives on the
> streaming device and then send whatever fragments of the message arrive
> on the device, until the message is complete. But I think that's not
> possible with the channel interface?
>
Yes, though so and would be great. Currently the RedChannelClient requires
each item to return a marshaller with all data and will send data from
the marshaller. RCC does not allow to "pause" with data not available.
> Cheers,
> Lukas
>
> > Christophe
>
Frediano
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