Convert Integer to int in Java

Question

I find a very weird situation when writing Java code:

Integer x = myit.next();
if ((int)x % 2 == 0) {

In which myit is an Iterator and x is an Integer. I just want to test whether x is an even number or not. But x % 2 == 0 does not work since eclipse says % not defined on Integer. Then I try to convert x to int by explicitly converting. Again, it warns me that not able to convert in this way.

Any reason why it happened and what is the right way to test if x is even ?

UPDATE: ANYWAY,I test it that the following code works, which means all of you guys are right.

    Integer x = 12;
    boolean y = ( (x % 2) == 0 );
    boolean z = ( (x.intValue() % 2) == 0 );

I think the problem I have before may be the context of the code. It is late night, I would update later if I find why would that thing happen.

Your code works fine when I test using Java 8. Are you possibly using an earlier version of Java> — CocoNess
– CocoNess, Commented Sep 17, 2014 at 7:02
@CocoNess It is the first time I met such a dilemma. I use Java 7 in this code. I find other code that use integer.intValue() works but my code does not work. — Kai
– Kai, Commented Sep 17, 2014 at 8:17

TheLostMind · Accepted Answer · 2014-09-17 06:57:17Z

5

Use :

if (x.intValue() % 2 == 0)

PS : if(x % 2==0) should also work because integer.intValue() should be called internally.

Byte code for :if(x % 2==0)

   11:  invokevirtual   #23; //Method java/lang/Integer.intValue:()I   --> line of interest
   14:  iconst_2
   15:  irem

edited Sep 17, 2014 at 6:57

answered Sep 17, 2014 at 6:48

TheLostMind

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2 Comments

Ruchira Gayan Ranaweera Over a year ago

This is not necessary. You can use % with Integer.

TheLostMind Over a year ago

@RuchiraGayanRanaweera - yes.. I was checking the byte code to reconfirm integer.intValue() was being called.

Ruchira Gayan Ranaweera · Accepted Answer · 2014-09-17 06:51:00Z

2

x % 2 == 0 does not work since eclipse says % not defined on Integer

This is not true. You can use % with Integer

take a look at this

Integer x = new Integer("6");
  if (x % 2 == 0) {
      System.out.println(x);
  }

Out put:

You should read about Integer in Java

answered Sep 17, 2014 at 6:51

Ruchira Gayan Ranaweera

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Comments

Ankur Singhal · Accepted Answer · 2014-09-17 06:50:31Z

1

public static void main(String[] args) {
        List<Integer> myList = new ArrayList<Integer>();
        myList.add(21);
        myList.add(22);
        myList.add(41);
        myList.add(2);

        Iterator<Integer> itr = myList.iterator();

        while (itr.hasNext()) {
            Integer x = itr.next();
            if (x % 2 == 0) {
                System.out.println("even");
            } else {
                System.out.println("odd");
            }
        }
    }

Output

odd
even
odd
even

answered Sep 17, 2014 at 6:50

Ankur Singhal

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Comments

Virginia Woolf · Accepted Answer · 2014-09-17 06:50:22Z

0

Use this:

if(((int)x)%2==0){

Note: one more (

answered Sep 17, 2014 at 6:50

Virginia Woolf

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