Exception Handling Java Program. You can give the input dynamically and can check it.
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Excepdemo {
/**
* @param args
*/
public static void main(String[] args) {
int ary[]={1,2,3};
try {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println(“Enter number”);
int a = Integer.parseInt(br.readLine());
System.out.println(“Enter number”);
int b = Integer.parseInt(br.readLine());
int c=a/b;
System.out.println(ary[10]);
}catch (ArithmeticException e) {
System.out.println(e);
}catch (ArrayIndexOutOfBoundsException e) {
System.out.println(e);
}
/* catch (NumberFormatException e) {
System.out.println(e);
} */
catch (IOException e) {
System.out.println(e);
}
catch(Exception e)
{
System.out.println(“u have done a mistake.”);
}
}
}