Equals Method.

[Abhi vijay]

When 2 primitive array references are checked using the equals() method, when are they considered equal??

  int[] i1= new int[3];
int[] i2= new int[3];
System.out.println(i1.equals(i2));  

Returns false. Is it because Array is an object and hence uses the == operator when using equals() method?

[Ruben Soto]

You are correct Abhi. Arrays are objects, and the equals method is not overridden to check for element-wise equality. You can use one of the utility methods in the Arrays utility class though. I think it is called deepEquals or something to that effect. You can check the API for it.

EDIT: Actually, deepEquals is just for reference variable arrays. For primitive arrays you would just use equals. There are a few overloaded versions of equals to compare pairs of arrays of every primitive type.

[Alpesh Rathod]

Well Keep one thing in mind that when its dealing with objects always use ==,because == always checks for location and not content whereas equals checks for content.
String is the only class which is not initialized wth new...and hence if the instance is of type String then we always go for equals method.
I hope i have cleared the things properly..if not rectify if possible...

[Ruben Soto]

Alpesh Rathod wrote:Well Keep one thing in mind that when its dealing with objects always use ==,because == always checks for location and not content whereas equals checks for content.
String is the only class which is not initialized wth new...and hence if the instance is of type String then we always go for equals method.
I hope i have cleared the things properly..if not rectify if possible...

Alpesh,

Actually, the equals method provided by Object internally uses the == operator, so unless a specific class overrides the method to check for logical equivalence, you will still be using ==. String is not the only class which is not initialized with new, the wrapper classes can be declared and initialized in one line using autoboxing without using the new keyword.

[Punit Singh]

Yes Ruben is right here.

For value -128 to 127, if you do not use new than Number wrappers do not create new object, they will give you cached object.

[Alpesh Rathod]

Hey Thanks for your Info..

[Abhi vijay]

String is not the only class which is not initialized with new, the wrapper classes can be declared and initialized in one line using autoboxing without using the new keyword.

Please, can someone elaborate?

[Ruben Soto]

Abhi vijay wrote:

String is not the only class which is not initialized with new, the wrapper classes can be declared and initialized in one line using autoboxing without using the new keyword.

Please, can someone elaborate?

Sure, do this:
Integer i = 1; //Declare and instantiate a new Integer object

[James Tharakan]

Abhi vijay wrote:

String is not the only class which is not initialized with new, the wrapper classes can be declared and initialized in one line using autoboxing without using the new keyword.

Please, can someone elaborate?

They are trying to say that...
its not just String object which can be created without using new keyword( ex:- String s = "james"; )
Even Wrapper objects can be created, ( Integer i = 2; )

[Abhi vijay]

Integer x1= new Integer(20);
	int x2=20;
	long l1=20l;
	System.out.println(x1==x2);
	System.out.println(x1==l1); // Line1
	System.out.println(l1==x2);
	System.out.println(x1.equals(x2));

	System.out.println(x1.equals(l1)); // Line 2   

I dont understand why line1 is giving TRUE and line2 FALSE???

[Ruben Soto]

Abhi, I think that line 1 is using autounboxing in the wrapper, and then applying the equality operator. The problem with line 2 is that the equals method of wrapper classes only returns true if the argument object is of the same type and holds the same value
as this object. In your case, you have an Integer and a Long, so equals must return false.

[Punit Singh]

Abhi vijay wrote:

Integer x1= new Integer(20);
	int x2=20;
	long l1=20l;
	System.out.println(x1==x2);
	System.out.println(x1==l1); // Line1
	System.out.println(l1==x2);
	System.out.println(x1.equals(x2));

	System.out.println(x1.equals(l1)); // Line 2   

I dont understand why line1 is giving TRUE and line2 FALSE???

System.out.println(x1==l1); // Line1

this line is using primitive comparison so, it is actually comparing 20=20.

System.out.println(x1.equals(l1)); // Line 2

this line calls Integer class's equals() method.
and in equals method thing is something like this:

equals(Object o){
if(!(o instanceof Integer))
return false;
....
....
}

so here l1 instanceof Integer fails.

[Abhi vijay]

Thanks Punit.

Can you tell me where can I find this information.

equals(Object o){
if(!(o instanceof Integer))
return false;
....
....
}

[Punit Singh]

you will find src.zip in your JAVA_HOME directory.

I am using eclipse.

In eclipse press ctrl and click on Integer word in your code, it will open a new Integer.class and you can see attach source button, just click on that button and give it the location of your src.zip file.

So java source code will be available in your eclipse, if you click on Integer word with ctrl button pressed you can see source code of Integer.class. You can use ctrl+o (o for onion) to find equals() method.

[Abhi vijay]

Thanks.