Equals Method

[Abhi vijay]

public class Invoke
{
public static void main(String[] args)
{
int i1=4000;
int i2=4000;

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

new Invoke().go(i1,i2);
}

public void go(Integer i1, Integer i2)
{
System.out.println(i1==i2);
}
}

The answer is true false.
But when i1=2; i2=2; then answer is true.
Why so?

[Mohit Mehta]

Abhi Vijay :

public class Invoke
{
public static void main(String[] args)
{
int i1=4000;
int i2=4000;
 
System.out.println(i1==i2);
 
new Invoke().go(i1,i2);
}
 
public void go(Integer i1, Integer i2)
{
System.out.println(i1==i2);
}
}

In case of go() method primitive int is auto-boxed to wrappered Integer.
It will create new objects, for new objects == will return false as their memory location would be different, whereas if you try .equals() method it will return true

[Pranav Patel]

Well I think,

i1=2, i2=2

: For ASCII value (-128 to 127), it will compare it as an

int

value (so gives true).

i1=4000, i2=4000

: For other

int

value, it will pass it as an

Integer

object as shown in the code (So gives false).

[Pranav Patel]

Sorry for the previous post, it is not readable..

i1=2, i2=2: For ASCII value (int -128 to 127), it will compare it as an int value (so gives true).
i1=4000, i2=4000: For other int value, it will pass it as an Integer object as shown in the code (So gives false).

Auto boxing for ASCII value for int works different here.

[Punit Singh]

Look code of Integer.valueOf() method in Integer class. You will get your answer.

If you call with i1=2, i2=2, then JVM will not going to create new Integer object. Why?

Here Integer.valueOf(2) method is get called not new Integer(2); And this method has already created a cache for Integer object of 256 size that will contain all Integer object starting from -128 to 127.
From -128 to 127, no new Integer objects are created, as they are already created and cached.

[Abhi vijay]

Thanks Punit and Pranav.