Boxing Doubt

[saloni jhanwar]

K&B Chapter3 page 246
Boxing

two instances of the following wrapper objects will always be = = when their primitive values are the same:

Boolean

Byte

Character from \u0000 to \u007f (7f is 127 in decimal)

Short and Integer from -128 to 127

class Auto 
{
	public static void main(String[] args) 
	{
		Integer i1=new Integer(2);
		Integer i2=new Integer(2);
				
		System.out.println(i1!=i2);           // why it is true ??
		System.out.println(i1.equals(i2));
		
		Integer i3=new Integer(1271);
		Integer i4=new Integer(1271);
				
		System.out.println(i3!=i4);
		System.out.println(i3.equals(i4));
	}
}

Output
C:\SCJP\Chapter3\AutoBoxing>java Auto
true
true
true
true

[Rameshwar Soni]

Hello please refer this thread Wrapper class

[saloni jhanwar]

Rameshwar Soni wrote:Hello please refer this thread Wrapper class

I didn't find answer of my question in your post. i1 and i2 are in given range then why they are not equal ?

[Rameshwar Soni]

saloni jhanwar wrote: i1 and i2 are in given range then why they are not equal ?

Yes i1 and i2 are in the given range but when you are using the new keyword to create Objects than irrespective of whether the number is in the given range or not Object will be created.

That means i1 and i2 will have different address. Hope this helps

[saloni jhanwar]

Rameshwar Soni wrote:

saloni jhanwar wrote: i1 and i2 are in given range then why they are not equal ?

Yes i1 and i2 are in the given range but when you are using the new keyword than irrespective of whether the number is in the given range or not Object will be created.

That means i1 and i2 will have different address. Hope this helps

two instances of the following wrapper objects will always be = = when their primitive values are the same:

Short and Integer from -128 to 127

I am not satisfied with your answer. Do you mean above written in K&B is wrong ?

[Rameshwar Soni]

saloni jhanwar wrote:

class Auto 
{
	public static void main(String[] args) 
	{
		Integer i1=new Integer(2);
		Integer i2=new Integer(2);
				
		System.out.println(i1!=i2);           // why it is true ??
		System.out.println(i1.equals(i2));
		
		Integer i3=new Integer(1271);
		Integer i4=new Integer(1271);
				
		System.out.println(i3!=i4);
		System.out.println(i3.equals(i4));
	}
}

As per your comment in your code your question is why i1 != 12 is true it must have been false.

As i said
Number 1 : When you are using "new" keyword Object will be created irrespective of any thing which means that created Object will have its own address.
Number 2 : == and != are used to check if 2 objects have same address or not and equals() method is used to check if 2 objects have same contents.

So when you wrote

Integer i1=new Integer(2);

i1 is created with address say for example 012fff
Now you wrote

Integer i2=new Integer(2);

i2 is created since you have used "new" keyword so i2 will have its own address say 013hhh

Now you wrote

System.out.println(i1!=i2);

Then definitely it will be true because i1 has a different address as compared to i2.

This is what i have understood from your comment in question. If i am wrong then i am sorry or be more specific

[saloni jhanwar]

I know about equal method so i am not comparing it with == .So question is that as you are right that address will be different of both wrapper class objects so they can't be same, then what case will be when two wrapper class objects will be == when they have same primitive value ? as written in book.Can you give me an example about it. thanks

[Anayonkar Shivalkar]

Hi saloni jhanwar,

What Rameshwar Soni is saying is correct. And what Kathy Sierra is saying is also correct.

Similar to String pool, there's something called as Integer cache in Java (Java SE 5 onwards).
e.g.

String str1 = new String("abc"); //creates a new String object
String str2 = "abc"; //looks for "abc" in String pool

Similarly, it works for Integer cache:

Integer i1 = new Integer(5); //creates a new Integer object

But when we say

Integer i2 = 5;

What we say is:

Integer i2 = Integer.valueOf(5);

Now, this Integer.valueOf method, it works quite interestingly. What it does is, it pics an Integer reference from cache for all Integers ranging from -128 to 127 (this cache is initialized by IntegerCache which is inner class of Integer class), and hence, when we say

Integer i3 = 5;
Integer i4 = 5;

It actually calls Integer.valueOf(5) two times, and since 5 falls in that range(-128 < 5 < 127), it will return exactly same object, and hence, i3 == i4 is true here.
But if we say

Integer i3 = 500;
Integer i4 = 500;

Then, cache won't be referred, and i3 == i4 would be false.

Hence, as Kathy said, Integer would be '==' when values are between -128 to 127, and as Rameshwar said, it won't happen if we use new keyword.

I hope this helps.

[Rameshwar Soni]

saloni jhanwar wrote:then what case will be when two wrapper class objects will be == when they have same primitive value ? as written in book.Can you give me an example about it.

As == is to check address therefore 2 Objects will be == when they have same address and not same primitive value (keeping the range -128 and 127 in mind).

Example :

 [b]case 1 :  Numbers within range and no use of "new" keyword[/b]
Integer i1 = 2;  // i1 created and has its own address...........note i haven't used "new" keyword here
Integer i2 = 2;   // i1 and i2  both have same address since value 2 was within the range -128 to 127 and also i haven't used "new" keyword
System.out.println (i1 == i2);     // true since == is used to check address and both i1 and i2 have same address
System.out.println (i1 != i2);      // false since != again checking address
System.out.println(i1.equals(i2));    // true since both have same content

[b]case 2 :  Numbers within range and  use of "new" keyword[/b]

Integer i3 =new Integer (3);  // i3 created and has its own address...........note i have used "new" keyword here
Integer i4 = new Integer (3);   // since i have used "new" keyword i3 and i4  both have different address even though they have same content
System.out.println (i3 == i4);     // false since == is used to check address and both i3 and i4 have different address
System.out.println (i3 != i4);      // true since != again checking address
System.out.println(i3.equals(i4));    // true since both have same content

[b]case 3 :  Numbers above range and no use of "new" keyword[/b]

Integer i5 = 200;  // i5 created and has its own address...........note i haven't used "new" keyword here
Integer i6 = 200;   // i6 created and has its own address since 200 is above range 
System.out.println (i5 == i6);     // false since == is used to check address and both i5 and i6 have different address
System.out.println (i5 != i6);      // true since != again checking address
System.out.println(i5.equals(i6));    // true since both have same content

[b]case 4 :  Numbers above range and  use of "new" keyword[/b]

Integer i7 = new Intger(300);  // i7 created and has its own address...........note i have used "new" keyword here
Integer i8 = new Integer(300;   // i8 created and has its own address since "new" keyword is used 
System.out.println (i7 == i8);     // false since == is used to check address and both i7 and i8 have different address
System.out.println (i7 != i8);      // true since != again checking address
System.out.println(i7.equals(i8));    // true since both have same content

[saloni jhanwar]

Rameshwar Soni wrote:

saloni jhanwar wrote:then what case will be when two wrapper class objects will be == when they have same primitive value ? as written in book.Can you give me an example about it.

As == is to check address therefore 2 Objects will be == when they have same address and not same primitive value (keeping the range -128 and 127 in mind).

Example :

// [b]case 1 :  Numbers within range and no use of "new" keyword[/b]
Integer i1 = 2;  // i1 created and has its own address...........note i haven't used "new" keyword here
Integer i2 = 2;   // i1 and i2  both have same address since value 2 was within the range -128 to 127 and also i haven't used "new" keyword
System.out.println (i1 == i2);     // true since == is used to check address and both i1 and i2 have same address
System.out.println (i1 != i2);      // false since != again checking address
System.out.println(i1.equals(i2));    // true since both have same content

//[b]case 2 :  Numbers within range and  use of "new" keyword[/b]

Integer i3 =new Integer (3);  // i3 created and has its own address...........note i have used "new" keyword here
Integer i4 = new Integer (3);   // since i have used "new" keyword i3 and i4  both have different address even though they have same content
System.out.println (i3 == i4);     // false since == is used to check address and both i3 and i4 have different address
System.out.println (i3 != i4);      // true since != again checking address
System.out.println(i3.equals(i4));    // true since both have same content

//[b]case 3 :  Numbers above range and no use of "new" keyword[/b]

Integer i5 = 200;  // i5 created and has its own address...........note i haven't used "new" keyword here
Integer i6 = 200;   // i6 created and has its own address since 200 is above range 
System.out.println (i5 == i6);     // false since == is used to check address and both i5 and i6 have different address
System.out.println (i5 != i6);      // true since != again checking address
System.out.println(i5.equals(i6));    // true since both have same content

//[b]case 4 :  Numbers above range and  use of "new" keyword[/b]

Integer i7 = new Intger(300);  // i7 created and has its own address...........note i have used "new" keyword here
Integer i8 = new Integer(300;   // i8 created and has its own address since "new" keyword is used 
System.out.println (i7 == i8);     // false since == is used to check address and both i7 and i8 have different address
System.out.println (i7 != i8);      // true since != again checking address
System.out.println(i7.equals(i8));    // true since both have same content

OMG thanks i will read it.

[saloni jhanwar]

Rameshwar Soni wrote:

saloni jhanwar wrote:then what case will be when two wrapper class objects will be == when they have same primitive value ? as written in book.Can you give me an example about it.

As == is to check address therefore 2 Objects will be == when they have same address and not same primitive value (keeping the range -128 and 127 in mind).

Example :

 [b]case 1 :  Numbers within range and no use of "new" keyword[/b]
Integer i1 = 2;  // i1 created and has its own address...........note i haven't used "new" keyword here
Integer i2 = 2;   // i1 and i2  both have same address since value 2 was within the range -128 to 127 and also i haven't used "new" keyword
System.out.println (i1 == i2);     // true since == is used to check address and both i1 and i2 have same address
System.out.println (i1 != i2);      // false since != again checking address
System.out.println(i1.equals(i2));    // true since both have same content

[b]case 2 :  Numbers within range and  use of "new" keyword[/b]

Integer i3 =new Integer (3);  // i3 created and has its own address...........note i have used "new" keyword here
Integer i4 = new Integer (3);   // since i have used "new" keyword i3 and i4  both have different address even though they have same content
System.out.println (i3 == i4);     // false since == is used to check address and both i3 and i4 have different address
System.out.println (i3 != i4);      // true since != again checking address
System.out.println(i3.equals(i4));    // true since both have same content

[b]case 3 :  Numbers above range and no use of "new" keyword[/b]

Integer i5 = 200;  // i5 created and has its own address...........note i haven't used "new" keyword here
Integer i6 = 200;   // i6 created and has its own address since 200 is above range 
System.out.println (i5 == i6);     // false since == is used to check address and both i5 and i6 have different address
System.out.println (i5 != i6);      // true since != again checking address
System.out.println(i5.equals(i6));    // true since both have same content

[b]case 4 :  Numbers above range and  use of "new" keyword[/b]

Integer i7 = new Intger(300);  // i7 created and has its own address...........note i have used "new" keyword here
Integer i8 = new Integer(300;   // i8 created and has its own address since "new" keyword is used 
System.out.println (i7 == i8);     // false since == is used to check address and both i7 and i8 have different address
System.out.println (i7 != i8);      // true since != again checking address
System.out.println(i7.equals(i8));    // true since both have same content

Thanks Rameshwar Soni it was very good example.

[Rameshwar Soni]

You're welcome