https://bugs.wireshark.org/bugzilla/show_bug.cgi?id=3957
Summary: RTP save payload lost packets handling
Product: Wireshark
Version: 1.2.1
Platform: Other
OS/Version: Windows XP
Status: NEW
Severity: Major
Priority: Medium
Component: Wireshark
AssignedTo: wireshark-bugs@xxxxxxxxxxxxx
ReportedBy: florentcapelle@xxxxxxxx
Build Information:
wireshark 1.2.1 (SVN Rev 29141)
Copyright 1998-2009 Gerald Combs <gerald@xxxxxxxxxxxxx> and contributors.
This is free software; see the source for copying conditions. There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.
Compiled with GTK+ 2.16.2, with GLib 2.20.3, with WinPcap (version unknown),
with libz 1.2.3, without POSIX capabilities, with libpcre 7.0, with SMI 0.4.8,
with c-ares 1.6.0, with Lua 5.1, with GnuTLS 2.8.1, with Gcrypt 1.4.4, with MIT
Kerberos, with GeoIP, with PortAudio V19-devel (built Jul 19 2009), with
AirPcap.
Running on Windows XP Service Pack 2, build 2600, with WinPcap version 4.1
beta5
(packet.dll version 4.1.0.1452), based on libpcap version 1.0.0, GnuTLS 2.8.1,
Gcrypt 1.4.4, without AirPcap.
Built using Microsoft Visual C++ 9.0 build 30729
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After an RTP stream Analysis, it is possible to save the payload of the RTP
stream.
There is something wrong in the way lost RTP packets are handled when saving
the payload. What woul be expected is that an empty payload (all zeros) is
inserted where a packet is missing. However this is not the case.
It seems that when X packets are missing between two valid packets, the number
of empty packets inserted in between is not X but X/2.
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