RMI binding to SAME port but DIFFERENT IP address on SAME host

[Alexander N. Spitzer]

If I have a machine with 3 virtual IP addresses (192.168.1.[5-7]), how
can I start 3 instances of the same RMI application (each started with
different properties/configs), each listening on the port 1234, but each
instance binds to a different ip address.

that is to say:
instance #1 binds to 192.168.1.5/port 1234
instance #2 binds to 192.168.1.6/port 1234
instance #3 binds to 192.168.1.7/port 1234

I guess I am looking for something like:
Registry registry = LocateRegistry.createRegistry(<IPADDR>,<PORT>);

I tried
System.setProperty("java.rmi.server.hostname","<IPADDR>");
but had no luck


current code looks something like this:

/******* server side *********/
System.getProperties().setProperty("java.rmi.server.hostname",serverip);
Registry registry = LocateRegistry.createRegistry(serverport);
registry.rebind("RMITest", this);

//serverip is the IP to bind to... server port is port 1234

/******* client side **********/
Registry registry=LocateRegistry.getRegistry(serverip,serverport);
RMITestInterface rmit=(RMITestInterface)(registry.lookup("RMITest"));
rmit.dosomethingcool();


second instance (even though serverip is different) complains
java.rmi.server.ExportException: Port already in use:1234; nested
exception is: java.net.BindException: Address already in use



These have to be seperate instances of the application for each
customer... I know that it is possible to run them on different ports,
but this is really not acceptable in a large clustered environment. this
would become a complete nightmare very quickly.


what am I missing?!?! I have googled high and low, and looked through
all my java books, but see no mention of such a scenario! anyone?

TIA!

-alex

 

[Alex Molochnikov]

[snipped]

I guess I am looking for something like:
Registry registry = LocateRegistry.createRegistry(<IPADDR>,<PORT>);

There is no method in LocateRegistry with this signature. You can create a
Registry object bound to a specific IP address using this form of the
method:

public static Registry createRegistry(int port, RMIClientSocketFactory csf,
RMIServerSocketFactory ssf) throws RemoteException

You will have to provide a class that implements the RMIServerSocketFactory
interface (the RMIClientSocketFactory parm can be null); this class will
create a ServerSocket that you will have to bind to the desired IP address.
Here is an example:

import java.rmi.server.*;
import java.io.*;
import java.net.*;
public class AnchorSocketFactory extends RMISocketFactory implements
Serializable
{
private InetAddress ipInterface = null;
public AnchorSocketFactory() {}
public AnchorSocketFactory(InetAddress ipInterface)
{
this.ipInterface = ipInterface;
}
public ServerSocket createServerSocket(int port)
{
ServerSocket serverSocket = null;
try
{
serverSocket = new ServerSocket(port, 50, ipInterface);
}
catch (Exception e)
{
System.out.println(e);
}
return (serverSocket);
}
public Socket createSocket(String dummy, int port) throws IOException
{
return (new Socket(ipInterface, port));
}
public boolean equals(Object that)
{
return (that != null && that.getClass() == this.getClass());
}
}

In your code do this:

AnchorSocketFactory sf = new AnchorSocketFactory(<your InetAddress>);
LocateRegistry.createRegistry(port, null, sf);

HTH

Alex Molochnikov
Gestalt Corporation
www.gestalt.com

 

[Sam]

If I have a machine with 3 virtual IP addresses (192.168.1.[5-7]), how
can I start 3 instances of the same RMI application (each started with
different properties/configs), each listening on the port 1234, but each
instance binds to a different ip address.

that is to say:
instance #1 binds to 192.168.1.5/port 1234
instance #2 binds to 192.168.1.6/port 1234
instance #3 binds to 192.168.1.7/port 1234

I guess I am looking for something like:
Registry registry = LocateRegistry.createRegistry(<IPADDR>,<PORT>);

I tried
System.setProperty("java.rmi.server.hostname","<IPADDR>");
but had no luck


current code looks something like this:

/******* server side *********/
System.getProperties().setProperty("java.rmi.server.hostname",serverip);
Registry registry = LocateRegistry.createRegistry(serverport);
registry.rebind("RMITest", this);

//serverip is the IP to bind to... server port is port 1234

/******* client side **********/
Registry registry=LocateRegistry.getRegistry(serverip,serverport);
RMITestInterface rmit=(RMITestInterface)(registry.lookup("RMITest"));
rmit.dosomethingcool();


second instance (even though serverip is different) complains
java.rmi.server.ExportException: Port already in use:1234; nested
exception is: java.net.BindException: Address already in use



These have to be seperate instances of the application for each
customer... I know that it is possible to run them on different ports,
but this is really not acceptable in a large clustered environment. this
would become a complete nightmare very quickly.


what am I missing?!?! I have googled high and low, and looked through
all my java books, but see no mention of such a scenario! anyone?

TIA!

-alex

Alex,

I doubt an end host keeps multiple ports for each active ip address.
Data from a nic card is passed up the protocol stack, and the source
ip address has been discarded by the time it reaches the port.

You might try a single RMI server listening on a single port, passing
data to a given instance of a jvm based on customer id.

Sam90

 

[Alexander N. Spitzer]

Alex,

I doubt an end host keeps multiple ports for each active ip address.
Data from a nic card is passed up the protocol stack, and the source
ip address has been discarded by the time it reaches the port.

You might try a single RMI server listening on a single port, passing
data to a given instance of a jvm based on customer id.

Sam90

Sam, in an ASP environment, it is a requirement to have seperate
instances for each client.

Alex Molochnikov gave me the answer I was looking for, which was:

There is no method in LocateRegistry with this signature. You can create
a Registry object bound to a specific IP address using this form of the
method:

public static Registry createRegistry(int port, RMIClientSocketFactory
csf,RMIServerSocketFactory ssf) throws RemoteException

Thanks Alex M. et al.!

 

[Esmond Pitt]

I doubt an end host keeps multiple ports for each active ip address.
Data from a nic card is passed up the protocol stack, and the source
ip address has been discarded by the time it reaches the port.

Sam, none of this guesswork is true.

 

[Sam]

Sam, in an ASP environment, it is a requirement to have seperate
instances for each client.

That actually is what I was proposing - but that's a moot point, since
you got Alex's correct answer.

Alex Molochnikov gave me the answer I was looking for, which was:

There is no method in LocateRegistry with this signature. You can create
a Registry object bound to a specific IP address using this form of the
method:

public static Registry createRegistry(int port, RMIClientSocketFactory
csf,RMIServerSocketFactory ssf) throws RemoteException

My bad - I did some speculation which I didn't check, and now see what
happened. Alex's answer didn't make to Google by the time I made my
post

Thanks Alex M. et al.!

Agreed.

Regards,
Sam90

 

[Sam]

Sam, none of this guesswork is true.

It isn't? How so?

Regards,
Sam90

 

[Esmond Pitt]

It isn't? How so?

Because what the OP described is perfectly possible, despite your doubts
and speculations. See ServerSocket(port,backlog,bindAddress). If you
don't know the answer please don't guess, it only confuses others.

 

[Sam]

Because what the OP described is perfectly possible, despite your doubts
and speculations.

I agree my speculation was incorrect, and I admit it and I'm not happy
about it. However, I did qualify it, that is I didn't say "this is the
gospel truth" or anything like that. I said "I doubt it", which
liberally translated means, "I'm shooting from the hip, beware".
Besides, nothing I said was actually false. I just want to clarify
that. Specifically, an end host doesn't track mutliple ports per ip -
does it? When data is passed up the protocol stack, the IP address is
discarded by the time it reaches

See ServerSocket(port,backlog,bindAddress).

I have, and I'm pretty sure I figured out where I went wrong.

If you
don't know the answer please don't guess, it only confuses others.

I did get sloppy, and I admit it. Oh well, no harm done, luckily
someone came along with the correct answer, actually even before I
posted my poor one. I'll be more careful next time.

Regards,
Sam90

 

[Esmond Pitt]

Sam

I don't really know what you mean by 'tracks multiple ports per IP',
which TCP certainly does otherwise nothing would work at all. Maybe you
mean 'multiple IP addresses per port'? in which case the answer is
certainly 'yes' as well, hence the ServerSocket constructor I referred
you to. The local IP address by which incoming data came in is not
reliable due to the 'weak end system model' described in the RFC, maybe
this is what you are referring to, but the IP address to which the
reading socket is bound is always available via Socket.getLocalAddress().

EJP

 

[Sam]

Sam

I don't really know what you mean by 'tracks multiple ports per IP',
which TCP certainly does otherwise nothing would work at all. Maybe you
mean 'multiple IP addresses per port'? in which case the answer is
certainly 'yes' as well, hence the ServerSocket constructor I referred
you to. The local IP address by which incoming data came in is not
reliable due to the 'weak end system model' described in the RFC, maybe
this is what you are referring to, but the IP address to which the
reading socket is bound is always available via Socket.getLocalAddress().

EJP

Esmond,

My thinking was: whatever OS routine is responsible for delivering
payload data to a listening application over a specific port doesn't
need to know what IP address the data came over, it only needs to
deliver it to the correct port. The IP address doesn't even exist at
the transport layer. So to my mind the only the way the requirement
could be achieved was to somehow set up separate ports with the same
port number for each listening interface, which seemed unlikely.

However, what I failed to account for is that Java Sockets speak with
IP at some level, because for example they can specify the remote IP
in the Socket class. So, from there it's not much of a jump to realize
it probably has a way to understand the local IP address, as well.

I'm still unclear as to the mechanics of it, though. I'm pretty sure
that on Windows Jave uses JNI to interact with the WinSock API to
handle it's network functionality. But, my real question is, what
exactly is a "port" anyway, except a numeric label which has a
constraint that it can't be re-used within a single IP address? In the
end, it may be that it's simply a question of processes - the Winsock
API or corresponding OS module probably tracks some internal table
which matches a port to a specific interface, and as long the
combination of the two is different from all other entries in its
table, no problem. In the end, what I thought was "unlikely" is
probably what actually occurs - the OS tracks a separate port for each
interface, or at least has the capability of doing so. (DISCLAIMER:
the preceding is pure speculation...)

Regards,
Sam90

 

[Esmond Pitt]

Sam

IP delivers data via sockets. That's what it does, it can't do anything
else. A socket is identified by its IP address, its port number, and its
protocol number. In the case of TCP, the socket is one end of a
connection whose other end is defined by *its* IP address and port
number. So, for the data to go anywhere at all IP has to know the local
address and port number *first*, i.e. it scans the table of sockets
somehow to find one with the target IP address/port/protocol no.
specified in the incoming IP packet. Then, the data is passed to that
socket, and the local IP address (& port) of any socket can be retrieved
via the Sockets API. This is nothing to do with Java, it all happens at
the IP layer. See W R Stevens 'TCP/IP Illustrated' for details. The only
time Java comes into it is by mapping the Sockets API into methods of
the Socket/ServerSocket classes, e.g. Socket.getLocalXXX().

EJP

 

[Roedy Green]

hat's what it does, it can't do anything
else. A socket is identified by its IP address, its port number, and its
protocol number.

I'm not sure if that is what you meant to imply, but there is no
"protocol number" field in a TCP/IP header.

see http://mindprod.com/jgloss/tcpip.html

The host computer simply knows which protocol it is hosting on each
port.

You can't have two different protocols hosted on the same port, e.g.
HTTP and FTP.

 

[Esmond Pitt]

Roedy

There is a protocol number field in an IP packet header, and I was
talking about IP at the time. Obviously in the case of TCP the protocol
number is the TCP protocol number, 6. BTW the statement that 'the host
computer simply knows which protocol it is hosting on each port' needs
further work. At the TCP level the host neither knows nor cares about
the protocol; at the IP level you *can* have several protocols on the
same port, e.g. ICMP, UDP, TCP.

EJP

 

[Sam]

Sam

IP delivers data via sockets. That's what it does, it can't do anything
else. A socket is identified by its IP address, its port number, and its
protocol number. In the case of TCP, the socket is one end of a
connection whose other end is defined by *its* IP address and port
number. So, for the data to go anywhere at all IP has to know the local
address and port number *first*, i.e. it scans the table of sockets
somehow to find one with the target IP address/port/protocol no.

Esmond,

If it does it this processing all in one go based on a single match of
ip address/port, then what's the point of having a layered protocol
stack? i.e, the link layer checks if the mac address is correct for
the current host, and if so passes the data upward to the IP module.
The IP module assures that the packet is intended for the current host
based on its IP address, and if so passes it up to the transport
module - regardless of port. Then the transport module looks for a
listening process on the destination port in its process/port table,
and if it finds one, then it passes the data to that process - if not
listening process is found, an ICMP error message is sent back to the
originating host. Isn't this the fundamental flow of a packet up the
tcp/ip protocol stack? If so, it seems to differ from the process you
describe, which seems to describe a single process for matching
ip/port.

Regards,
Sam90

specified in the incoming IP packet. Then, the data is passed to that
socket, and the local IP address (& port) of any socket can be retrieved
via the Sockets API.

This is nothing to do with Java, it all happens at

 

[Esmond Pitt]

If it does it this processing all in one go based on a single match of
ip address/port,

Sam

If you want to know how TCP works please read the reference I mentioned
before. I didn't say say anything about everything happening in one go,
this is more guesswork on your part. Obviously the data is passed from
the NIC to IP to TCP and obviously lots of processing happens on the
way, but an IP socket is *defined* by its protocol number, an IP
address, and a port number, and these socket attributes are available
all the way up the stack and indeed to the application.

EJP

 

[Sam]

Sam

If you want to know how TCP works please read the reference I mentioned
before. I didn't say say anything about everything happening in one go,
this is more guesswork on your part. Obviously the data is passed from
the NIC to IP to TCP and obviously lots of processing happens on the
way, but an IP socket is *defined* by its protocol number, an IP
address, and a port number, and these socket attributes are available
all the way up the stack and indeed to the application.

EJP

EJP,

I'm not convinced what you've been saying is entirely accurate. For
example, you previously mentioned that data doesn't go anywhere unless
there is a socket available. You also state that socket attributes are
available all the way up the stack.

I've just reviewed pp. 19-21 of Steven's TCP Illustrated, Volume II.
He describes the progress of a UDP packet up the protocol stack, from
the hardware interrupt which launches the nic driver, to the passing
of the data by the driver to the IP layer, which in turn passes it to
the UDP input routine.

The first mention of sockets is made in the description of the UDP
input routine. It uses the port on the UDP header to locate a PCB
Control block, and when it finds a match "This will be the PCB created
by our call to socket, and the inp_socket member of this PCB points to
the corresponding socket structure, allowing the received data to be
received by the correct socket". My reading this has the data going
as least as far as the UDP (transport) layer, and no socket available
to the stack until then.

This appears to contradict your assertion about the packets not going
anywhere without a socket - it gets as far as the transport layer -
and about the socket being available throughout the protocol stack -
it's only looked up at the transport layer.

The whole point of the layered architecture is to prevent one layer
from having to contend with data that belongs in another layer, and
that's been cited as one of the main reason for TCP/IP's great
success.

Regards,
Sam90

 

[Tony Morris]

This thread is somewhat difficult to follow, so I'll assume the question is
the typical, "how do I have an RMI server bind to a specific network
interface?"

The answer is you use java.rmi.registry.LocateRegistry, create a
RMIServerSocketFactory that is bound to a specific interface using
java.net.NetworkInterface.

If my assumption is incorrect, please ignore.

 

[Esmond Pitt]

Sam

OK, if UDP or TCP or ICMP or whatever transport layer matches the socket
to the inbound IP packet, instead of IP, the local IP address *still*
doesn't get 'thrown away as the data goes up the protocol stack',
because it must still be there in the IP packet for TCP or UDP to do
this matching, and it must also available be in the socket descriptor to
be matched. Can we agree on that?

This also means that TCP, UDP, ICMP &c do indeed deal with IP data, &
just goes to show that TCP/IP implementation layering is not as clean as
you might like to think. Certainly IP doesn't have to deal with TCP
data, but the other way round does happen (no worse than a derived class
using methods from its base class really).

Of course Stevens vol II only describes one implementation. I don't know
that a 'cleaner' implementation where IP despatches packets to sockets
is infeasible either ... What *would* be infeasible is the
implementation you seem to keep imagining where TCP can't access the IP
header or the local IP address: otherwise the OP's situation wouldn't
work, and it does; also Socket.getLocalAddress() == getsockname()
couldn't work either.

Anyway basta cosi.

EJP

 

[Esmond Pitt]

This thread is somewhat difficult to follow, so I'll assume the question is
the typical, "how do I have an RMI server bind to a specific network
interface?"

The answer is you use java.rmi.registry.LocateRegistry, create a
RMIServerSocketFactory that is bound to a specific interface using
java.net.NetworkInterface.

If my assumption is incorrect, please ignore.

That works for the registry itself, but for specific RMI servers you
also have to provide *them* with a ServerSocketFactory, probably the
same one, and you have to use the appropriate IP address or interface
name when binding them to the registry.