How Wrapper classes works

Question

Integer x = 400;
Integer y = x;
x++;

At this point x is 401 and y is 400.But I think both are referring to the same object and both should be 401. I don't know how this is happening.please help.

Title seems misleading, I think you need to see this http://stackoverflow.com/questions/4712798/how-to-use-references-in-java or http://stackoverflow.com/questions/40480/is-java-pass-by-reference — Pranav Singh, Jun 25 '13 at 06:45
http://stackoverflow.com/questions/13098143/java-integer-constant-pool — Santosh, Jun 25 '13 at 06:56

score 7 · Answer 1 · answered Jun 25 '13 at 06:41

7

Because x++ is effectively the same as x = x + 1. x + 1 is a different Integer object, and x is updated to refer to this other object.

answered Jun 25 '13 at 06:41

Oliver Charlesworth

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+1 due to auto-boxing, it might not create a new Integer, but it will be a different Integer. e.g. `-XX:+AggressiveOpts` can increase the Integer cache to 10K or 20K – Peter Lawrey Jun 25 '13 at 06:42
pls look at my answer, it proves from different perfective. – Ahmet Karakaya Jun 25 '13 at 06:57

score 2 · Answer 2 · answered Jun 25 '13 at 06:56

Opcode says everything.

package wrapperInteger;

public class WrapperTest {


public static void main(String[] args) {

    Integer x =400;
    Integer y=x;
    x++;
    y--;


}

}

OPcode:

If you realized that x corresponds #16 and y corresponds #22. So it proves that both variables are pointing different objects.

Compiled from "WrapperTest.java"
public class wrapperInteger.WrapperTest extends java.lang.Object{
public wrapperInteger.WrapperTest();
  Code:
   0:   aload_0
   1:   invokespecial   #8; //Method java/lang/Object."<init>":()V
   4:   return

public static void main(java.lang.String[]);
  Code:
   0:   sipush  400
   3:   invokestatic    #16; //Method java/lang/Integer.valueOf:(I)Ljava/lang/Integer;
   6:   astore_1
   7:   aload_1
   8:   astore_2
   9:   aload_1
   10:  invokevirtual   #22; //Method java/lang/Integer.intValue:()I
   13:  iconst_1
   14:  iadd
   15:  invokestatic    #16; //Method java/lang/Integer.valueOf:(I)Ljava/lang/Integer;
   18:  astore_1
   19:  aload_2
   20:  invokevirtual   #22; //Method java/lang/Integer.intValue:()I
   23:  iconst_1
   24:  isub
   25:  invokestatic    #16; //Method java/lang/Integer.valueOf:(I)Ljava/lang/Integer;
   28:  astore_2
   29:  return
}

Thanks...can you please tel me more about opcode..what does it describe ..how to get this code — sp_user123, Jun 25 '13 at 14:39
>javap -c WrapperTest.class prints java compiled codes. Opcode is like a machine language which is executed by JVM. JVM executes this code. — Ahmet Karakaya, Jun 25 '13 at 19:49

score 1 · Answer 3 · answered Jun 25 '13 at 06:44

1

All wrapper classes in java are immutable.So a new instance of the object creates each time. Of course carrying different value for each instance.

answered Jun 25 '13 at 06:44

Suresh Atta

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score 0 · Answer 4 · answered Jun 25 '13 at 06:43

0

Integer is immutable . So when you are copying some values to it it is creating a new instance of the object

answered Jun 25 '13 at 06:43

stinepike

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score 0 · Answer 5 · answered Jun 25 '13 at 07:05

0

Here In this image, when you increment value of X=X+1 then it will create new reference rather then pointing to same one.So after incrementing value of X, X will point to 401.And Y will point to 400 as Y is not incremented.

Check This

answered Jun 25 '13 at 07:05

Manish Doshi

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How Wrapper classes works

5 Answers5