Hi Frank,
In the default constructor,
public DefaultUdpTransportMapping() throws IOException {
super(new UdpAddress(InetAddress.getLocalHost(), 0));
socket = new DatagramSocket(udpAddress.getPort());
}
the UdpAddress is initialized with localhost as you mentioned. But only the
port is passed to DatagramSocket and thus the ip address it binds to is a wild
card ipaddress.
But in renewSocetAfterException(), Socket is initialized using both ip and port
which causes the mismatch.
DatagramSocket s = new DatagramSocket(udpAddress.getPort(),
udpAddress.getInetAddress());
Thanks,
Prema
-----Original Message-----
From: SNMP4J [mailto:[email protected]] On Behalf Of Frank Fock
Sent: Tuesday, July 22, 2014 4:57 PM
To: [email protected]
Subject: Re: [SNMP4J] renewSocketAfterException() binds to a specific address
Hi Prema,
The renewSocketAdterException(..) method of the DefaultUdpTransportMapping uses
*exactly* the same UDP address as provided in the constructor. If you provided
"0.0.0.0", it will be used again.
You wrote that you are using the default constructor. That one is *not* using
"0.0.0.0", instead it uses the localhost IP address. It depends on your
operating system and Java virtual machine which IP that is on a multi-home
system.
Best regards,
Frank
Am 22.07.2014 20:18, schrieb Prema Upot:
> Hi,
>
> We initialize DefaultUdpTransportMapping using the default constructor and
> this makes sure the listening port binds to "0.0.0.0" address enabling
> sending and receiving UDP packets on any interface. But in
> renewSocketAfterException() the new socket is created using port and
> ipaddress as parameters and this causes a problems in a PC that is part of
> multiple networks.
> For example, we have a PC in 192.168.144.0 and 192.168.30.0 networks. After
> socket exception, the new socket that is created binds to
> "192.168.144.<ip>:<port>" instead of "0.0.0.0:<port>" and because of this we
> cannot communicate with 30.0 network any more.
>
> Thanks,
> Prema
>
>
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