How to specify a local port but it should not be bound

Hi all,
i am developing a client app that connects to a server...
for example i have the constuctor of a socket that connects to a server..the server requires the ipaddress and local port of the client for security reasons...so my socket constructor looks like this client = new Socket (remote ipaddress, remote port, local ipaddress, local port) ..this works fine...but the problem is that the server sends messages to the local port.....how do i make sure that java does not bind..the local port..i mean for it to be freed..cos it seems java is binding it.
Thanks.

there is a Server (Not a java program) that requires
clients to connect to it using the following kind of
connection below:
a java client connects to the server with this socket
constructor
Socket ss = new Socket (remote ip, remote port, local ip, local port);This is your first problem. There is no such kind of server. The server will know the client's local port regardless of which Socket constructor the client uses, as someone else pointed out. Leave out the last two parameters.
now the
server sends responses to the local port specified in
the constructor of the socket.or the local port assigned by the system if it wasn't specified, as it almost always shouldn't be.
I need to create a listener on the localport
so i can read the information sent in from the serverAHA!!!!!!! NO YOU DO NOT. You need to execute a read on the client socket. That's what it's for. It already represents an established connection between a client local port and the server.
..creating
this ServerSocket ss = new ServerSocket
(localport); in jave generates an "Address
Already in Use Exception".As expected. It is.
now what i asked was "how
can i unbind the local port form the socket so java
will not tell me it is already in use when creating a
listener on it in ServerSocket".You don't want to do this as explained before. FYI you could close the Socket to release the port, but you will then discover that the server sends fail because it tries to send to the established connection.
I hope this is clear now.Clear as mud. What is clear is that you do indeed need to read up on TCP/IP as recommended by several other posters. Your grasp is very slender.

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