Why == resulting in true , in Long object

Question

In this piece of code (story * 2) == tail is getting True

and false for distance + 1 != tail.

== checks for reference , as Long is immutable , it will false for two different objects,

Here The value story * 2 is getting equal in reference to tail , but they are two different objects and not a compile time constant for pooling.

   public class Test2 
{
         public static void main(String [] args) {

              Long tail = 2000L;
              Long distance = 1999L;
              Long story = 1000L;

                  System.out.println(tail > distance);

                  System.out.println((story * 2) == tail);

              if((tail > distance) ^ ((story * 2) == tail))
                  System.out.print("1");

              System.out.println(distance + 1 != tail);
              System.out.println((story * 2) == distance);

              if((distance + 1 != tail) ^ ((story * 2) == distance))
              System.out.print("2");

}

I checked here , but no explanations for this.

Integer does but not Long , for Integer -128 to 127 are cached but not for long and surely not in this question because values are greater than 127 — anshulkatta, Jun 24 '13 at 05:46
@Pathashu note that the values in the question are larger than 127, I suppose that caching is not the reason here — Andreas Fester, Jun 24 '13 at 05:47
Check out this answer: - http://stackoverflow.com/questions/10149959/using-operator-in-java-to-compare-wrapper-objects — Marek, Jun 24 '13 at 05:50
@Marek no that is not the answer , already checked it , in that integer values are less than 128 thats why it is getting true — anshulkatta, Jun 24 '13 at 05:51

NPE · Answer 1 · 2013-06-24T05:55:24.107

7

When you perform arithmetic operations on wrapped primitives (such as Long), they are automatically unboxed into raw primitives (e.g. long).

Consider the following:

(story * 2) == tail

First, story is auto-unboxed into a long, and is multiplied by two. To compare the resulting long to the Long on the right-hand side, the latter is also auto-unboxed.

There is no comparison of references here.

The following code demonstrates this:

public static void main(String[] args) {
    Long tail = 2000L;
    Long story = 1000L;
    System.out.println((story * 2) == tail);          // prints true
    System.out.println(new Long(story * 2) == tail);  // prints false
}

edited Jun 24 '13 at 05:55

answered Jun 24 '13 at 05:48

NPE

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(story * 2) results in long , ok , then == is done between long and Long , why will for == Long will be auto-boxed – anshulkatta Jun 24 '13 at 05:52
because one of the items of compare is primitive. In such case JVM always unbox the type to primitive. – Damian Leszczyński - Vash Jun 24 '13 at 06:11

score 4 · Accepted Answer · answered Jun 24 '13 at 05:49

4

I beleive it is due to the auto-unboxing when you do (story * 2) resulting in primitive value 2000L. And when you compare it against tail which also hold 2000L value, hence the result is true. Check here the x==y rule when one item is primitive.

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Source: http://www.javapractices.com/topic/TopicAction.do?Id=197

answered Jun 24 '13 at 05:49

Juned Ahsan

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why Long is converted long and not reverse when == happens ?? – anshulkatta Jun 24 '13 at 06:01
got it thanks !! read on http://www.javapractices.com/topic/TopicAction.do?Id=197 – anshulkatta Jun 24 '13 at 06:03
1

Long is converted to long when you do (story * 2) due to auto-unboxing. – Juned Ahsan Jun 24 '13 at 06:03
yes , and tail is converted to long , as == happens between primitives only , if one is primitive then object is changed to primitive – anshulkatta Jun 24 '13 at 06:10

score 0 · Answer 3 · answered Jun 24 '13 at 05:49

Multiplication, addition and other operations can be carried out on primitives only...

When you carry out those operations all the boxed primitives will be unboxed and treated as primitive types.

So in your example == is checking long equality not object equality.

Why == resulting in true , in Long object

3 Answers3