Java

nogono · #1 (**permalink**) 07-29-2007, 10:52 AM

What exactly do you mean by this syntax?

if(this.running!=running)
{

}

HelloWorld · #2 (**permalink**) 07-29-2007, 11:01 AM

Quote:

Originally Posted by nogono

What exactly do you mean by this syntax?

if(this.running!=running)
{

}

That means you have both instance variable of running and local variable of running. this.running refers to the instance of the running variable (outside of your method), while running is refers to the local variable.

For example:

PHP Code:

  private int hello = 1;
public void test() {
     int hello = 2;
     System.out.println(this.hello); // printing 1
     System.out.println(hello); // printing 2
}

Hope that helps

nogono · #3 (**permalink**) 07-29-2007, 11:34 AM

I understand your concept of running...thaks a lot for that..
but i am confused about

if(this.running!=running)
{

}

if this.running is false and running is false
will it enter the loop.....
if yes how?
if not why not?

basically i am trying to get the condition in if loop....

HelloWorld · #4 (**permalink**) 07-29-2007, 11:39 AM

no, that means that if this.running is NOT EQUAL to running (assuming that they're not object. Meaning that they're just a primitive type of variables, then you may use that comparison

PHP Code:

  public void test() {
     int one = 1;
     int two = 2;
     int test = 1;
     if (one != two) {
          System.out.println("test"); 
          // will be executed since they are NOT equal
     }
     if (one == two) {
          System.out.println("test2");
          // won't be executed cuz they are NOT equal
     }
     if (one == test) {
          System.out.println("test3");
          // will be executed because they are equal
     }
}

would that helps you

nogono · #5 (**permalink**) 07-29-2007, 11:55 AM

Thanks a lot for previous help....u are too good to explain anyone...

socket client=null;
serversocket listen_socket;
client=listen_socket.accept();

is it true that only port is getting assigned to socket client.b'coz serversocket only have port....

HelloWorld · #6 (**permalink**) 07-29-2007, 12:00 PM

I'm not sure with what you're trying to say, but about networking, here's what happened:

ServerSocket (in the server) will have all of the clients' socket, and each time ServerSocket accept a connection to a client, that client should get a listener (which you will create with multi threading), that way there will be more than one client can connect simultaneously.

port is not assigned to the client, port is opened from the server, and the client will connect to that specific port.

If you're not sure with multi threading, feel free to see my previous tutorial about multi threading, I think I made several of them

(they're all in Java forum)

nogono · #7 (**permalink**) 07-29-2007, 01:12 PM

public static final int variable=100;

When we are intializing variable as static...is there any necessity to declare it final......
will be fine if public static int variable=100;

HelloWorld · #8 (**permalink**) 07-29-2007, 01:14 PM

Quote:

Originally Posted by nogono

public static final int variable=100;

When we are intializing variable as static...is there any necessity to declare it final......
will be fine if public static int variable=100;

It's depending on your needs, sometimes we want that variable to be final, so that it can't be changed

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