plz look at this code

i get the error [B]undefined label: newHost[\B]

but not a similar error for label intError

Code:

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import java.io.*;
import java.net.*;

public class Client {
        public static void main (String args []) throws IOException {
                BufferedReader keyboard = new BufferedReader (new InputStreamReader (System.in));
                               
                                 
                [I]newHost:[\I] System.out.print ("connect to IP: ");
                String host = keyboard.readLine();
               
                int port= 0;
               
               
                [I]intError: [\I]try {
                        port = Math.abs (Integer.parseInt (keyboard.readLine()));
                }
               
                catch (NumberFormatException e) {
                        System.out.println ("plz enter a valid port num");
                        break intError;
                }
               
               
                try {Socket connection = new Socket (host, port);}
                catch (UnknownHostException e){
                        System.out.println ("Host can not be found\n");
                        break newHost;
                }
                               
               
        }//end main
}//end Client

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That's interesting code. I can't help too much, except to parrot an excerpt from Bruce Eckel's very good book (available online) "Thinking in Java". It looks pertinent tho:


A label is an identifier followed by a colon, like this:


label1:
The only place a label is useful in Java is right before an iteration statement. And that means right before—it does no good to put any other statement between the label and the iteration. And the sole reason to put a label before an iteration is if you’re going to nest another iteration or a switch inside it. That’s because the break and continue keywords will normally interrupt only the current loop, but when used with a label they’ll interrupt the loops up to where the label exists:


label1:
outer-iteration {
inner-iteration {
//...
break; // 1
//...
continue; // 2
//...
continue label1; // 3
//...
break label1; // 4
}
}
In case 1, the break breaks out of the inner iteration and you end up in the outer iteration. In case 2, the continue moves back to the beginning of the inner iteration. But in case 3, the continue label1 breaks out of the inner iteration and the outer iteration, all the way back to label1. Then it does in fact continue the iteration, but starting at the outer iteration. In case 4, the break label1 also breaks all the way out to label1, but it does not re-enter the iteration. It actually does break out of both iterations.

I see what you are trying to do. It looks like if an invalid host or port is detected then you want an exception to be thrown and program execution to continue at the given label. If this is your first draft i would just scrap it and start over with a new program structure. I do this all of the time.

Try using the crab claw ie..... for(;;) :p I perfer lobster myself but.... The code below loops until either an "x" or an "X" is entered. If a MalformedURLException is thrown the method just propagates back to it's caller which is contained within the endless for(;;) loop.

Code:

---

  import java.io.*;
  import java.net.*;

  public class URLReader{
  public static void main(String[] args){
 
  System.out.println("Welcome to the URL Reader Program \n");
  System.out.println("Please Enter a URL.Note~* Url must be in http:// format!\n");
  System.out.println("To exit the program just press x or X\n");
 
  try{
  BufferedReader buff = new BufferedReader(new InputStreamReader(System.in));
 
  for(;;){
  String url = buff.readLine();
    if(url.equals("x") || url.equals("X")){
      System.exit(0);
    }
  startReader(url);
     
  }
  }catch(IOException e){System.out.println(e);}
  } 

  public static void startReader(String url){
  try{
    URL u = new URL(url);
    InputStream in = u.openStream();
    StreamCopier.copy(in,System.out);
    in.close();
    }
    catch(MalformedURLException e){System.err.println(e);
    }
    catch(IOException e){System.err.println(e);} 
  }
}

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