How's it work?

Write here if you have problems with your Java source code

Moderator: Board moderators

Post Reply
Roby
Experienced poster
Posts: 101
Joined: Wed May 04, 2005 4:33 pm
Location: Tangerang, Banten, Indonesia
Contact:

How's it work?

Post by Roby »

Can someone explain to me how Java examine this code:

Code: Select all

public class Everything
{
 public static void main( String [] args )
 {
    Integer a = new Integer(5);
    Integer b = new Integer(5);
    if ( a == b )
       System.out.println( "same" );
 }
}
Why doesn't it worked? Thanx in advance :)
Darko
Guru
Posts: 580
Joined: Fri Nov 11, 2005 9:34 am
Location: Calgary, Canada

Post by Darko »

Code: Select all

a==b
evaluates if both a and b are referencing the same object. In this case they are not. You probably want to use

Code: Select all

a.equals(b)
But, in Java 1.5 you can do this:

Code: Select all

Integer a = 5;
Integer b = 5;
if(a==b)...
Same thing has always been the case with Strings - if you declare them using literals like in the last example, a==b will evaluate to true - because of the way JVM stores the literals. I guess now it does it the same way with all the primitives wrapped into an object.
Roby
Experienced poster
Posts: 101
Joined: Wed May 04, 2005 4:33 pm
Location: Tangerang, Banten, Indonesia
Contact:

Post by Roby »

So, since java 1.5, the primitive which wrapped into object can be compared using that method...

OK, thanx Darco, just a bit confusing :D
impulser
New poster
Posts: 4
Joined: Wed Jun 28, 2006 4:43 am

Post by impulser »

hey how do the JVM handle the code for other java versions . Since, i think we won't upgrade the JVM versions .. as it is ported into the OS..

Hey Darko , can you be specific about the way JVM handles it..

Thanks ... :D
Darko
Guru
Posts: 580
Joined: Fri Nov 11, 2005 9:34 am
Location: Calgary, Canada

Post by Darko »

Well, I'm not 100% sure, but this is how I understand it:

Code: Select all

Integer a = new Integer(5); // JVM allocates memory for Integer object and a points to it
Integer b = new Integer(5); // same as above

if(a==b)... // JVM checks if both a and b point to the same object (they don't)

if(a.equals(b))... // JVM checks if two Integers referenced (did I say "pointer" up there?:)) by a and b are equal according to the equals() implementation - if it weren't overridden in Integer class, it would have the same effect as a==b, but in this case it translates to
if(a.intValue() == b.intValue())...
Now, in the case of Strings:

Code: Select all

String a = new String("abc");
String b = new String("abc");
// this is exactly the same as with Integer example above

a==b - false
a.equals(b) - true
Now the fun part, if you do it this way (you could always do this with Strings):

Code: Select all

String a = "abc"; //JVM checks its pool of literal strings for "abc", if it is there, a points to it, if not, JVM adds "abc" to the pool and a points to it
String b = "abc"; // same as the above

a==b is true! (so is a.equals(b))
I hope that's what you wanted to know. :)
Post Reply

Return to “Java”