Handling Consecutive NumberFormat Exceptions

Question

I have written a Binary Search Program for which I am asking the user to enter the size and the elements of the array.
If the user enters the size say 5
Now he can enter up to a max of 5 max numbers to populate the array.
If the user enters an alphabet it should not be accepted and it would throw an exception and its caught properly for the first time and the user is asked to enter the number again. Now again if the user enters an alphabet my program is not robust enuf to throw an error to the user and continue the user to enter a number from that point. How do i handle such exceptional conditions.
import java.util.Arrays;
import javax.swing.JOptionPane;
public class BinarySearch {
     public static void main(String[] args) {
          String n = JOptionPane.showInputDialog(null, "Enter the size of the array");
          int size = Integer.parseInt(n);
          int a[] = new int[size];
          int i = 0;
          while(i < size){
               String arrayn = JOptionPane.showInputDialog(null, "Enter the "+(i+1)+" th element of the array");
               try{
                    a[i] = Integer.parseInt(arrayn);
               }catch (Exception e){
                    arrayn = JOptionPane.showInputDialog(null, "Enter the "+(i+1)+" th element of the array");
                    a[i] = Integer.parseInt(arrayn);
               i++;
          System.out.print("The elements of the array are : ");
          for(int j = 0; j < size; j++){
               System.out.print(a[j]+"  ");     
          Arrays.sort(a);               //Sorting the numbers for Binary Search
          System.out.println();
          System.out.print("The Sorted elements of the array are : ");
          for(int j = 0; j < a.length; j++){
               System.out.print(a[j]+"  ");     
          String key = JOptionPane.showInputDialog(null, "Enter the element to search for from the array");
          int searchKey = Integer.parseInt(key);
          int position = binarySearch(a, searchKey);
          System.out.println();
          if(position == -1){
               System.out.println("The element could not be found");
          }else{
               System.out.println("The element found at "+position+"th position of the array");
     public static int binarySearch(int a[], int searchKey){
          int low, high, middle;
          int index = -1;
          low = 0;
          high = a.length - 1;
          middle = (low + high + 1) / 2;
          do{
               if(searchKey == a[middle]){
                    return middle;
               }else if(searchKey < a[middle]){
                    high = middle -1;
               }else{
                    low = middle + 1;
               middle = (low + high + 1) / 2;
          }while(low <=high && (index == -1));
          return index;
}Message was edited by:
hemanthjava

Answer 1

Hi,
I made the changes. I am getting NumberFormat Error for the size even after making a change.
int k = 0;
          while(k < 1){
               try{
                    n = JOptionPane.showInputDialog(null, "Enter the size of the array");
                    k++;
               }catch (Exception e) {
                    System.out.println("Please enter a number for size");
                    continue;
Exception in thread "main" java.lang.NumberFormatException: For input string: "j"
     at java.lang.NumberFormatException.forInputString(Unknown Source)
     at java.lang.Integer.parseInt(Unknown Source)
     at java.lang.Integer.parseInt(Unknown Source)
     at BinarySearch.main(BinarySearch.java:19)
Message was edited by:
hemanthjava