Concerning "else" statement in for-loop when traversing through array

Question

I'm trying to solve a problem where I need to find all the even numbers only. I need to input 5 numbers, and if none of the numbers are even I want to print Even number not found in array. So my problem is when I iterate through the for-loop, my code prints Even number not found in array. It prints for each non-even number, which is not what its suppose to do obviously. I need some kind of hint. This is not homework btw, it is a problem found on Programmr.com. Here is my code:

 import java.util.Scanner;

 public class ArrayEven {

   public static void main(String args[]) {
     @SuppressWarnings("resource")
     Scanner scanner = new Scanner(System.in);
     int x, arr[] = new int[5];

     for (int i = 0; i < arr.length; i++) {
       arr[i] = scanner.nextInt();
       if (i == 4)
         break;

    }

     for (int i = 0; i < arr.length; i++) {

       x = arr[i] % 2;

       if (x == 0) {

         System.out.println(arr[i]);

       }

       else if (x != 0) {  //this is obviously wrong. Do I need another for-loop for this?

         System.out.println("Even number not found in array.");

       }

     }

   }

 }

Not a bug · Accepted Answer · 2014-01-21 09:50:03Z

3

You can use a boolean here, Initialize boolean variable with false if you found any even than set the boolean as true and check that boolean in your if condiion.

Example :

boolean isAvailble = false;
...
// Some code
...
for (int i = 0; i < arr.length; i++) {
       x = arr[i] % 2;
       if (x == 0) {
         System.out.println(arr[i]);
         isAvailble = true;
       }
}

if (! isAvailable) {  
     System.out.println("Even number not found in array.");
}

answered Jan 21, 2014 at 9:50

Not a bug

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2 Comments

deathshot Over a year ago

You guys are awesome. It works and actually makes alot of sense now.

Not a bug Over a year ago

Pleasure that we can help someone...:)

Pratik Shelar · Accepted Answer · 2014-01-21 09:52:33Z

1

public static void main(String args[]) {
 @SuppressWarnings("resource")
 Scanner scanner = new Scanner(System.in);
 int x, arr[] = new int[5];

 for (int i = 0; i < arr.length; i++) {
   arr[i] = scanner.nextInt();
   if (i == 4)
     break;

}
boolean evenFound = false; 

 for (int i = 0; i < arr.length; i++) {

   x = arr[i] % 2;

   if (x == 0) {

     System.out.println(arr[i]);
     evenFound = true; 

   }

 }
 if(!evenFound){
     System.out.println("Not found");

  }

}

answered Jan 21, 2014 at 9:52

Pratik Shelar

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