Thanks for the answer. I also think I don't get your point :D I try to explain my understand of your point with a little graphic.
Timeline: sec 0---------------1----------2--------------------------------------------------------------3-----4--------------------------------------------------------------------------------------5-----6-------------------------> Sender: StartRecord0(); StopRecord0();StartSend0();StartRecord1(); FinishSend0();StopRecord1();StartSend1();StartRecord2(); FinishSend1();StopRecord2();StartSend2();StartRecord3(); Receiver: StartReceive0(); FinishReceive0();StartPlay0();StartReceive1(); StopPlay0():FinishReceive1(); StartPlay1();StartReceive2(); In my opinion that's the way you mean. But as you see, the recorded video from second 0 to 2 will be played on the receiver side from second 4 to 6. Maybe it can be optimized to play the received video-stream directly after the StartReceive, but so the video is played from second 2 to 4. That means there is a delay of at least 2 seconds. Am I right? On Saturday, March 17, 2012 8:20:18 AM UTC+1, Daniel Drozdzewski wrote: > > > > On Friday, 16 March 2012, paul <paulelsne...@googlemail.com> wrote: > > Thank you for than answer, but that isn't a solution for the problem. > Doing it that way, leads to a delay depending on the recording time to that > file. Also there would be an interruption of the video stream. > > I don't think you got the idea... Recording would be happening to fileB > while previously recorded fileA would be sent through the socket at the > same time. The only interuption in the video stream would be due to > stopping of the recording, passing the recorded file to the sender, > reseting of the recorder and setting it up with the new file. All that > should take under a second. > > HTH > > > I want to use the captured and encoded video for videoconferencing, so > any kind of delay is not acceptable. > >> ..do you mean storing it in the buffer and creating a delay/jitter? > > It seem to me that the video encoder works in that way. I don't want a > delay. > > > > Am Freitag, 16. März 2012 17:54:29 UTC+1 schrieb Daniel Drozdzewski: > >> > >> On 16 March 2012 15:22, paul <paulelsne...@googlemail.com> wrote: > >> > Then it would write out the data when the stop function is called. > >> > >> ..do you mean storing it in the buffer and creating a delay/jitter? > >> > >> You will have to look at double buffering using 2 files ... while one > >> is being recorded to, the other will be sent through the socket. Once > >> writing is finished, you stop the recorder, point it at new file and > >> restart it, while taking just finished file and sending it. Sent > >> files can be discarded or left for the sake of having a local copy. > >> > >> -- > >> Daniel Drozdzewski > > > > -- > > You received this message because you are subscribed to the Google > > Groups "Android Developers" group. > > To post to this group, send email to android-developers@googlegroups.com > > To unsubscribe from this group, send email to > > android-developers+unsubscr...@googlegroups.com > > For more options, visit this group at > > http://groups.google.com/group/android-developers?hl=en > > -- > Daniel Drozdzewski > -- You received this message because you are subscribed to the Google Groups "Android Developers" group. To post to this group, send email to android-developers@googlegroups.com To unsubscribe from this group, send email to android-developers+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/android-developers?hl=en