GStreamer-CRITICAL **: gst_segment_to_running_time: assertion `segment->format == format' failed
Tim-Philipp Müller
t.i.m at zen.co.uk
Mon Sep 23 15:34:28 PDT 2013
On Mon, 2013-09-23 at 15:44 +0200, Peter Maersk-Moller wrote:
> I have a pipeline that reports:
>
> GStreamer-CRITICAL **: gst_segment_to_running_time: assertion
> `segment->format == format' failed
>
>
> I tried to search mailing list and the web, but found no obvious ideas
> to what it means. What does it mean? And how does one goes around
> mitigating this? It should be said that the pipeline plays. The
> version used is 1.1.90
>
>
>
> The pipeline is shown below. I notice that the CRITICAL message is no
> longer present, if I remove the pipeline elements AFTER the 'filesink'
> element. So quite likely something between decodebin and shmsink
> provokes the critical message, but I'm not really sure what to look
> for.
>
> CAPS='application/x-rtp, media=(string)video, clock-rate=(int)90000,
> encoding-name=(string)H264, sprop-parameter-sets=(string)"Z0JAKLtAKALb
> +AokAAADAAQAAAMAwYEAALcbAAtx73vheEQjUA\=\=\,aM44gA\=\=",
> payload=(int)96'
> MIXERFORMAT=' video/x-raw, format=(string)BGRA,
> pixel-aspect-ratio=(fraction)1/1, interlace-mode=(string)progressive,
> width=(int)1280, height=(int)720'
>
> /usr/local/bin/gst-launch-1.0 -v \
> fdsrc fd=0 !\
> $CAPS !\
> rtph264depay !\
> h264parse !\
> tee name=t1 !\
> queue !\
> rtph264pay !\
> udpsink host=192.168.1.1 port=19044 t1. !\
> queue !\
> mpegtsmux !\
> tsparse !\
> filesink location=somefile.ts t1. !\
> queue !\
> decodebin !\
> videoscale !\
> videoconvert !\
>
> $MIXERFORMAT !\
> shmsink socket-path=/tmp/feed3-control-pipe shm-size=22118400
> wait-for-connection=0 sync=false
>
>
> A producing pipeline could be something similar to
>
>
> gst-launch -q videotestsrc is-live=true do-timestamp=true !\
>
> $MIXERFORMAT !\
>
> x264enc !\
>
> h264parse !\
>
> rtph264pay !\
>
> fdsink fd=3 sync=true 3>&1 1>&2
It complains because fdsrc does not send a segment event in TIME format.
What kind of socket is fd=3 ? Sending rtp packets over a stream
connection won't work usually, because that won't maintain the
packetisation.
Cheers
-Tim
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