java exception handling

please any body help me out?what is the role of compiler and JVM when an exception arises in a java program?

//may be this document could help you
-->classAdd without Exception
{ {
public static void main( String[] arguments)
{
intnumber1;
intnumber2;
intsum;
number1 = Integer.parseInt(arguments[0]);
number2 = Integer.parseInt(arguments[1]);
sum = number1 + number2;
System.out.printf( "Sum is %d\n", sum );
}
}

-->classAdd with Exception

public class Add
{ {
public static void main( String[] arguments)
{
intnumber1;
intnumber2;
intsum;
try {
number1 = Integer.parseInt(arguments[0]);
number2= IntegerparseInt(arguments[1]); number2 = Integer.parseInt(arguments[1]);
sum = number1 + number2;
System.out.printf( "Sum is %d\n", sum );
} catch (Exception e) {
System.out.printf( "You did something terrible!\n\n" );
}
}
}

//May be this documant could help
public class Add
{ {
public static void main( String[] arguments)
{
intnumber1;
intnumber2;
intsum;
try {
number1 = Integer.parseInt(arguments[0]);
number2= IntegerparseInt(arguments[1]);
sum = number1 + number2;
System.out.printf( "Sum is %d\n", sum );
} catch (Exception e) {
System.out.printf( "You did something terrible!\n\n" );
}
}
}

uzair what are you trying to say?

the compiler wont compile source code if there is a checked exception that is not handled or declared. while you can, the compiler does not require you to handle runtime exceptions.

This is only an example.

uzair ahmed wrote:This is only an example.

you need to fix the example as that will never compile and will be very confusing to them.

import java.util.Scanner;
public class Square
{
public static void main( String[] args)
{
try
{
Scanner input = new Scanner( System.in);
int number;
int square;
System.out.print("EnteranInteger:");
number = input.nextInt(); //check for integer and then for character
square = number * number;
System.out.printf( "Square of %d is %d\n", number, square );
}
catch (Exception e)
{
System.out.printf( "You did something terrible!\n\n" );
} finally
{
System.out.printf( "Do Something regardless of exception!\n");
}
}
}

uzair ahmed wrote://may be this document could help you . . .

I don’t think it does help at all, I am afraid.

Apart from your posting unformatted, uncompilable code (read this), your exception handling is very poor. Using Exception, when you can predict what sort of Exceptions you might suffer, and that non‑specific error message, and not good style at all.

Nitya Sundar: welcome to the Ranch and sorry you have been given such poor advice. For exceptions, start reading here in the Java Tutorials. That should answer your question; if it doesn’t be sure to ask again

Thanks to you all.

You’re welcome

I hope you found what you needed.