Help ! Newbie !

[Jeremy Chee]

Hi there,
I'm a student from Singapore learning Java Programming
Currently facing some problems on my assignment.

how do i display a variable of value 0000

i tried the following

short oNum = 0000;
System.out.println(+ oNum );

it would display as 0
Any way to display all the 4 zeros.

Many Thank in advance.

[Jaime M. Tovar]

Try using DecimalFormat. It is an util object that helps to display numbers in the format you need. Look for java api in the web so you can take a look on the documentation.

[Jeremy Chee]

got it.
Thanks for the help.

[Dana Bothner-By]

String value = "0000";
System.out.println(value);

[Burkhard Hassel]

package tests;
 
import java.text.*;
 
class Main {
public static void main(String [] args) {
String [B]value[/B] = "0000";
System.out.println(value);
                               //       :wink: 
System.out.println("[B]Dana was cheating[/B]" + 
" up to this point, but");
DecimalFormat form = new DecimalFormat([B]value[/B]);
for (int i=0; i<5; i++)
System.out.println([B]form.format(i)[/B]);
System.out.println(form.format(5)+"...");
}
}

[Dana Bothner-By]

Cheating? What rule did I violate? Ockham's razor gives me a clean shave!

[marc weber]

Originally posted by Dana Bothner-By:
Cheating? What rule did I violate? ...

I think Bunkhard is simply pointing out (with a friendly ) that assigning a String literal is one thing, but formatting a primitive value is another.

[Jeremy Chee]

Haha..
You guys are like the pros man.
I get stuck sometimes.
Heres another one.

public class ABCInput
{
public static void main(String[]args)
{
char c = 'x';
System.out.print("Enter a character:");
c =(char)System.in.read();
System.in.read();
System.in.read();
System.out.println("You entered["+c+"]");
}
}

with this, i get an error message on line 7,8 and 9
unreported exception java.io.IOException; must be caught or declared to be thrown

Btw, i'm using Jcreator.
The codes may seem simple, coz i'm learning the basics in school.

[fred rosenberger]

You should read the Java API. Your line 7 calls System.in.read(). So, look up the System class (it's listed in that bottom left pane in the API). There, you can find that the 'in' is an "InputStream". So, look up that object. it has a method called "read".

that method throws an IOException.

Have you read up on error handling and exceptions? There is a quick, easy fix that will haunt you later if you continue using it, and then there is a slightly more complicated but better fix, but that requires some understanding of exceptions.

[Jeremy Chee]

Sorry, but i'm just learning the basic, so we're not taught all the way.
Is there any way to modify the code to make it work ?
That was from a sample code, but i couldnt get it to work.
I'm trying to implement it in one of my projects.

[fred rosenberger]

Java has a concept called "exceptions". Basically, they are used to help keep your program from dying when something goes horribly awry. When "Something Bad Happens", a special object is created, and you have to deal with it.

The problem you are seeing is that the method you want to use, the read() method, is defined as one that can throw an IOException. You can't change the fact that this method MIGHT throw that exception, when "Something Bad Happens".

If you want to use this method, you have to tell Java that you know this exception might be thrown, and then you have to deal with it somehow.

the simplest way would be to wrap your three calls to read() in a try/catch block, and then go from there. something like this:

try {
  c =(char)System.in.read();
  System.in.read();
  System.in.read();
}
catch (IOException e) {
  System.out.println("Something Bad Happened!!! OH NO!!!!");
}

this should let you get past the compilation error you are seeing (although it might need to be tweaked a little - this is just off the cuff). Then, when "Something Bad Happens", you should see that line print, and you will know why things don't go as planned.

You should read up on error handling, try/catch blocks, and exceptions (they're all related).