Error:Exception in thread "main" java.lang.OutOfMemoryError: Java heap space

Question

My program is as follows:

import java.util.*;

class evenNumber {
    ArrayList<Integer> arrL=new ArrayList<Integer>();

    ArrayList<Integer> saveEvenNumber(int N) {
        if(N<2)
            System.out.println("N should be greater than 2");
        else 
            for(int i=1;i<N;i++)
            {
                while(i%2==0)
                {
                    arrL.add(i);
                }
            }

        return arrL;
    }

    void printEvenNumber() {
        Iterator<Integer> tr=arrL.iterator();
        while(tr.hasNext())
            {
                System.out.print(tr.next());
                System.out.print("*2, ");
            }
    }
}


public class First {
    public static void main(String args[]) {
        evenNumber eN=new evenNumber();
        eN.saveEvenNumber(13);
        eN.printEvenNumber();
    }
}

I am getting the following error:

Exception in thread "main" java.lang.OutOfMemoryError: Java heap space
    at java.util.Arrays.copyOf(Unknown Source)
    at java.util.Arrays.copyOf(Unknown Source)
    at java.util.ArrayList.grow(Unknown Source)
    at java.util.ArrayList.ensureCapacityInternal(Unknown Source)
    at java.util.ArrayList.add(Unknown Source)
    at list.evenNumber.saveEvenNumber(First.java:15)
    at list.First.main(First.java:35)`

I have changed the size in Run > Run configuration > Arguments as mentioned in other posts on the same error but then also I am getting the same error.

What should I do?

You keep adding elements to your list in while(i%2==0) { arrL.add(i); }. Independent of your memory settings, this will run out of memory sooner or later. — Andreas Fester
– Andreas Fester, Commented Jul 21, 2016 at 7:18
In addition to Andreas' comment I guess you want to use if(i%2==0) instead. — Thomas
– Thomas, Commented Jul 21, 2016 at 7:19
Your way of iterating in printEvenNumber is one way to do it, but a simple for loop would be more readable: for (Integer i : arrL) { System.out.print(i); System.out.print("*2, "); } would do the same. — glglgl
– glglgl, Commented Jul 21, 2016 at 7:34

Eugene · Accepted Answer · 2016-07-22 11:26:31Z

6

When i == 2, the while loop will be executed forever and that's the reason why java.lang.OutOfMemoryError is thrown.

Add a break; after arrL.add(i);.

As Thomas commented, use if statement is more suitable here.

BTW, refer to Naming Conventions for java naming. For your case, the first letter of class name should be capitalized. Use EvenNumber instead of evenNumber.

edited Jul 22, 2016 at 11:26

answered Jul 21, 2016 at 7:19

Eugene

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1 Comment

Thomas Over a year ago

An if-statement would be the better solution here.

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