search for: packetcounter

Hi guys I succesfully got my encoder and decoder working after much hassles, but when I use the same code in another project, I get these following errors: Error ---> Invalid sideband mode encountered (1st sideband): 7 Error ---> Invalid sideband mode encountered (1st sideband): 7 Error ---> Invalid sideband mode encountered (1st sideband): 6 Error ---> More than two

Hi guys I have tried compiling this attached code, I made all the buffers 320, there is no trace of a 160 buffer, but I get a " SpeexEncoder requires 320 samples to process a Frame, not 160" error. Maybe there's something I'm missing, here's my code: import java.io.IOException; import java.io.FileOutputStream; import java.io.File; import

Okay, how do I drop a changeset/patchset/tag for you folks from SVN? At this point, I have written three examples of how to use the basics of the ogg streaming and decoding in Tremor. I heartily welcome any suggestions, improvements and corrections that you can point out in the code. The examples required me to make some small modifications to the main tremor library. However, the changes

Hi everyone, I just changed the meaning of the timestamp in the jitter buffer. If you don't know what I'm talking about, then you're not affected. If you do use the jitter buffer, than you will need to change your code accordingly, so instead of bumping the timestamp by 20 (ms) for each frame, you'll have to increase by 160 (samples) for narrowband or 320 for wideband. The new

...= 32768, that happens after 102 puts... I would say that an int is 32 bits :-) Actually, RTP defines the timestamp as a 32-bit value. Now, what happens when it overflows (3 days for narrowband), I don't know what the RFC says about it. > In my current > implementation I do "((long)packetCounter * 20) % 32760", but speex doesn't > like me starting over at zero, and resets the buffer (every 32 seconds)...How > to do it right? Why do you insist on having considering only 16 bits? Jean-Marc

...ld be closer to what RTP does (and not > have fractional value for sample rates like 44.1 kHz). Hi, what happens if this number flows over? It is just a "int", so it might reach its limits at 2^15 = 32768, that happens after 102 puts...In my current implementation I do "((long)packetCounter * 20) % 32760", but speex doesn't like me starting over at zero, and resets the buffer (every 32 seconds)...How to do it right? Peter -- The absent ones are always at fault. -------------- next part -------------- A non-text attachment was scrubbed... Name: not available Type: applicat...

...puts... > > I would say that an int is 32 bits :-) Actually, RTP defines the > timestamp as a 32-bit value. Now, what happens when it overflows (3 days > for narrowband), I don't know what the RFC says about it. > > > In my current > > implementation I do "((long)packetCounter * 20) % 32760", but speex > > doesn't like me starting over at zero, and resets the buffer (every 32 > > seconds)...How to do it right? > > Why do you insist on having considering only 16 bits? My C book taught me an int is only guaranteed to be 16bits or bigger, and si...