Integer unboxing and == operator question

[randy pausch]

Page 284 of the K&B book states

In order to save memory, two instances of the
following wrapper objects (created through boxing), will always be == when their
primitive values are the same:

Short and Integer from -128 to 127

Why does the same rule NOT apply to values over 127 ? What is so special about the -128 to 127 (byte) range?

Sample Code:

 

Integer i1 = 1000;
Integer i2 = 1000;
if(i1==i2)
{
System.out.println("==");
}
if(i1!=i2)
{
System.out.println("!=");
}

prints "!="

while

 
Integer i1 = 20;
Integer i2 = 20;
if(i1==i2)
{
System.out.println("==");
}
if(i1!=i2)
{
System.out.println("!=");
}

prints "=="

[Ankit Garg]

Just search in the forum for "integer pool" or "integer 127" and you'll get a result like this...

[randy pausch]

Sorry...I shud have searched before asking.

But guessing the right search query would've been a challenge though.

Thanks Ankit.

[Ankit Garg]

Payel please start a new topic for your problem by clicking .

Also when you post the code in the new topic, please Use Code Tags...

[Payel Bera]

Ankit Garg wrote:Payel please start a new topic for your problem by clicking .

Also when you post the code in the new topic, please Use Code Tags...

Hi ankit,

The topic was a continuation of the same query so does not need to be included as a separate topic/ new topic.
Regarding code tag will keep that in mind in the next post, for sure.
BTW can you/ anyone else provide the explanation, thanks

[Ankit Garg]

Payel, you can edit your message using button and then add code tags to it.

Coming to the question, what part of the output did you not understand??

[Tim McGuire]

In order to save memory, two instances of the
following wrapper objects (created through boxing), will always be == when their
primitive values are the same:

Short and Integer from -128 to 127

"created through boxing" is the key.

Integer.valueOf() returns an int . This int is autoboxed into an Integer, so the "Created through boxing" requirement is met with this one.

new Integer(10) : no boxing here. rule does not apply.

[Abimaran Kugathasan]

    
public class MyTest {

	public static void main(String args[])	{
		Integer i1 = new Integer (1000);
		Integer i2 = new Integer (1000);
		Integer i3 = new Integer (10);
		Integer i4 = new Integer (10);
		Integer i5 = Integer.valueOf("900");
		Integer i6 = Integer.valueOf("900");
		Integer i7 = Integer.valueOf("90");
		Integer i8 = Integer.valueOf("90");

		if(i1 == i2) System.out.println("i1 == i2");
		else System.out.println("i1 != i2");
		if(i3 == i4) System.out.println("i3 == i4");
		else System.out.println("i3 != i4");
		if(i5 == i6) System.out.println("i5 == i6");
		else System.out.println("i5 != i6");
		if(i7 == i8) System.out.println("i7 == i8");
		else System.out.println("i7 != i8");


	}
}

This your code....

The below syntax is only creates a new wrapper object on the Integer Literal Pool. If the value within -128 to 127, they will be referred by two reference variable, but actually, there will be only one object.

Integer i =100;
Integer j = 100;

Others will create objects on the heap as like any 'Other' objects.

So the == method, since it check the bit patterns of the object reference, will give the output like above....

[Payel Bera]

Thank you Tim and Abimaran for your prompt responses. Highly appreciate your explanations.