CODEWITHSUNDEEP
javasockets
Baruch
Baruch

Reputation: 21518

State of a new Java socket

When I declare in Java

Socket s = new Socket((String)null, 12345);

Does this actually open a socket and use system and network resources, or is that deferred until I attach an input/output buffer? I would like to create a Socket object at the start of my program that is all set up to connect to the server, and just open/close it as necessary, instead of having to pass an address and port around (it seems cleaner), but not if it means the port will be open the entire time.

EDIT
It seems from the answers that this will not work like I wanted. How can I create a closed socket that is all set up with address and only needs to connect?

Upvotes: 3

Views: 343

Answers (4)

Karthik Kumar Viswanathan
Karthik Kumar Viswanathan

Reputation: 1806

http://docs.oracle.com/javase/6/docs/api/java/net/Socket.html#Socket(java.net.InetAddress,%20int) <- it depends on the constructor you use. For the constructor you have specified, it connects.

Upvotes: 3

user207421
user207421

Reputation: 310957

Every constructor of Socket creates an underlying socket, which uses system resources, and all but the no-args constructor connect it as well, which uses network resources. There is no such operation as 'attach[ing] an input/output buffer' to a Socket.

Upvotes: 0

Polygnome
Polygnome

Reputation: 7795

For your edit: You will have to set up your own class that holds all setup information and can then be opened later. Maybe yüu'll just store the data in there and make a method that returns a socket. It's up to you, there's plenty ways to do that. But make sure all sockets were correctly closed at the end ;)

Upvotes: 0

William
William

Reputation: 1131

According to http://docs.oracle.com/javase/7/docs/api/java/net/Socket.html#Socket(java.lang.String,int) ,the way you are initializing your object it will be connected.

Upvotes: 2

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