Solved

Sockets

Posted on 2004-03-26
19
300 Views
Last Modified: 2008-02-01
why does socket.getLocalAddress() put a slash in front of the adress
ex. /192.168.1.100
0
Comment
Question by:tricks801
  • 9
  • 8
  • 2
19 Comments
 
LVL 92

Accepted Solution

by:
objects earned 500 total points
ID: 10691523
cause thats the format toString() uses.

If u don't want it try:

getLocalAddress().getHostName();
or
getLocalAddress().getHostAddress();
0
 
LVL 30

Expert Comment

by:GrandSchtroumpf
ID: 10691526
because there is no defined protocol.
0
 
LVL 30

Expert Comment

by:GrandSchtroumpf
ID: 10691536
with a defined protcol such as ftp or http, you would get
ftp:/192.168.1.100 or http:/192.168.1.100
so without the protocol, you still keep the "/".
0
Courses: Start Training Online With Pros, Today

Brush up on the basics or master the advanced techniques required to earn essential industry certifications, with Courses. Enroll in a course and start learning today. Training topics range from Android App Dev to the Xen Virtualization Platform.

 
LVL 92

Expert Comment

by:objects
ID: 10691547
> because there is no defined protocol.

An IP address doesn't have a protocol does it?
0
 
LVL 92

Expert Comment

by:objects
ID: 10691563
The slash is actual used to seperate hostname and address. But in your case the name is unknown so it is blank.
0
 
LVL 30

Expert Comment

by:GrandSchtroumpf
ID: 10691577
note that when you define a protocol and a port,
you use something like this http:/192.168.1.100:80
you can see that the separator is ":"
so, again, if you dont specify neither the protocol nor the port, you just get "/192.168.1.100"
0
 
LVL 30

Expert Comment

by:GrandSchtroumpf
ID: 10691597
> An IP address doesn't have a protocol does it?
i agree, but we're not talking about an ip address here, we're talking about an instance of SocketAddress...  and its toString() method as you mentioned before
0
 
LVL 92

Expert Comment

by:objects
ID: 10691600
> note that when you define a protocol and a port,

An IP address does not have a protocol or a port.

> you use something like this http:/192.168.1.100:80

That is not an IP address, 192.168.1.100 is the IP address

0
 
LVL 30

Expert Comment

by:GrandSchtroumpf
ID: 10691608
i was just explaining why the toString method leaves the slash.
0
 
LVL 92

Expert Comment

by:objects
ID: 10691615
> but we're not talking about an ip address here

I think we actually are

> we're talking about an instance of SocketAddress

Even if we were the same applies.
0
 
LVL 30

Expert Comment

by:GrandSchtroumpf
ID: 10691655
oops, it's not an instance of SocketAddress but an instance of java.net.InetAddress.
SocketAddress is for ServeSocket.
javadoc says about InetAddress:
"This class represents an Internet Protocol (IP) address."
which talks about "Protocol" as well as "(IP) address".
0
 
LVL 92

Expert Comment

by:objects
ID: 10691696
Yes the *Internet* Protocol :) Thats what IP stands for.
0
 
LVL 30

Expert Comment

by:GrandSchtroumpf
ID: 10691716
> The slash is actual used to seperate hostname and address. But in your case the name is unknown so it is blank.
ok, that's fine with me.
0
 
LVL 30

Expert Comment

by:GrandSchtroumpf
ID: 10691727
i was reading the wrong javadoc page.  :-)
0
 
LVL 14

Expert Comment

by:Tommy Braas
ID: 10691750
There are several layers of protocols. The Internet Protocol (as in IP) sits on the lower levels of the protocol stack and handles routing of packets. The TCP part of TCP/IP handles packet ordering, and connections for connection based communication.

An IPv4 address is x.y.z.w, where each x,y,z and w can hold values same as an unsigned byte. A port, however is local "routing" information, dependent on a protocol sitting on top of TCP/IP. You need both to make a connection to another computer over TCP/IP.

http, which is another protocol, but an application layer protocol, sits on top of TCP/IP, but is separate from TCP/IP.
0
 
LVL 30

Expert Comment

by:GrandSchtroumpf
ID: 10691813
> http, which is another protocol, but an application layer protocol, sits on top of TCP/IP, but is separate from TCP/IP.
i agree, http has nothing to do with this question.
sorry for all this mess i created.
0
 
LVL 14

Expert Comment

by:Tommy Braas
ID: 10691879
No worries mate  ;-)
0
 
LVL 92

Expert Comment

by:objects
ID: 10691888
> sorry for all this mess i created.

sok :)
0
 
LVL 92

Expert Comment

by:objects
ID: 10749684
0

Featured Post

Courses: Start Training Online With Pros, Today

Brush up on the basics or master the advanced techniques required to earn essential industry certifications, with Courses. Enroll in a course and start learning today. Training topics range from Android App Dev to the Xen Virtualization Platform.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

Java contains several comparison operators (e.g., <, <=, >, >=, ==, !=) that allow you to compare primitive values. However, these operators cannot be used to compare the contents of objects. Interface Comparable is used to allow objects of a cl…
Java had always been an easily readable and understandable language.  Some relatively recent changes in the language seem to be changing this pretty fast, and anyone that had not seen any Java code for the last 5 years will possibly have issues unde…
Viewers learn about the “while” loop and how to utilize it correctly in Java. Additionally, viewers begin exploring how to include conditional statements within a while loop and avoid an endless loop. Define While Loop: Basic Example: Explanatio…
This tutorial covers a step-by-step guide to install VisualVM launcher in eclipse.

815 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question

Need Help in Real-Time?

Connect with top rated Experts

7 Experts available now in Live!

Get 1:1 Help Now