Java RMI registry port change issue

Question

You can create a registry at a given port as follows.

import java.rmi.registry.LocateRegistry;
import java.rmi.registry.Registry;
...
Registry reg;
...
reg = LocateRegistry.getRegistry(port);

That's fine. But how do I change the port number of the Registry (reg) when needed, after initializing it for the first time?.

I've tried to recreate, reg = LocateRegistry.getRegistry(port), but that gives some exceptions..

Please show the exceptions – Yair Zaslavsky Jan 10 '13 at 18:58 — Yair Zaslavsky, Jan 10 '13 at 18:58

score 0 · Answer 1 · edited May 23 '17 at 10:24

0

If this port is already in use by rmi you can't close it:

It is a known limitation of the current RMI implementation that it never closes server (passive/listening) TCP port bindings for the lifetime of a virtual machine.> Blockquote

source: http://bugs.sun.com/view_bug.do?bug_id=4508962

Maybe this can help you:

But please post exception.

edited May 23 '17 at 10:24

Community

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answered Jan 10 '13 at 19:08

ollo

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Oh please. Read your source again. It *was* a known limitation, *in 2001.* That bug was fixed nearly ten years ago. – user207421 Jan 10 '13 at 20:14

user207421 · Answer 2 · 2013-01-12T02:38:15.410

You can create a Registry at a given port as follows

No you can't. That code just constructs a stub referencing a Registry at that port, which may not even be running. The way to do what you said is with LocateRegistry.createRegistry().

How do I change the port number of the Registry

Unexport the object returned by createRegistry() and call createRegistry() with a different port number.

Why you would want to do such a strange thing is another matter. What's the purpose?

I've tried to recreate

No you haven't, see above. All that does is create a stub pointing to another port. It doesn't create a Registry. If there isn't a Registry running on that port the stub will fail when used.

Java RMI registry port change issue

2 Answers2