I think using 5060 when nothing is specified is the correct thing to do and
we should change the documentation.

Peter

Peter Higginson
Newport Networks Ltd,
http://www.newport-networks.com/

-----Original Message-----
From: [EMAIL PROTECTED]
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Sent: 21 June 2007 17:28
To: [EMAIL PROTECTED]
Subject: [Sipp-users] [ sipp-Bugs-1741058 ] 5060-5120 always used
innon-random order if -p not specifie

Bugs item #1741058, was opened at 2007-06-21 12:28
Message generated for change (Tracker Item Submitted) made by Item Submitter
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https://sourceforge.net/tracker/?func=detail&atid=637564&aid=1741058&group_i
d=104305

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Category: Execution
Group: None
Status: Open
Resolution: None
Priority: 5
Private: No
Submitted By: Larry Low (llow)
Assigned to: Nobody/Anonymous (nobody)
Summary: 5060-5120 always used in non-random order if -p not specifie

Initial Comment:
Documenation states
   -p               : Set the local port number.  Default is a random free
port
                      chosen by the system.

but this is not true as it starts at 5060 and increases by 1 till it finds a
free port.

Suggest changing
    unsigned short l_port;
    for(l_port = DEFAULT_PORT;
        l_port < (DEFAULT_PORT + 60);
        l_port++) {

to something more like (10000 to 29999 range)

    unsigned short l_port;
    srand ( time(NULL) );
    unsigned short first_port = (rand()%20000) + 10000;
    for(l_port = first_port;
        l_port < (first_port + 60);
        l_port++) {

that or update the documentation and add a parameter of such.

The reasoning behind this is that I use Sipp to run scenarios and alarm if
the scenario fails (SIP Monitoring).  I want messages not to get confused
between calls in case there is some problem so I would prefer a random port
each time and since specifying a port does that is already in use will cause
a return of 255 immediately.

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