int vs Integer comparison Java [duplicate]

Question

class datatype1 {      public static void main(String args[])     {     int i1 = 1;     Integer i2 = 1;     Integer i3 = new Integer(1);      System.out.println("i1 == i2"+(i1==i2));     System.out.println("i1 == i3"+(i1==i3));     System.out.println("i2 == i3"+(i2==i3)); }  } 

Output

i1 == i2true i1 == i3true i2 == i3false 

Can someone explain why I get false when comparing i2 and i3 ?

Answer 1

i1 == i2 

results in un-boxing and a regular int comparison is done. (see first point in JLS 5.6.2)

i2 == i3  

results in reference comparsion. Remember, i2 and i3 are two different objects. (see JLS 15.21.3)

Answer 2

Integer i2 = 1; 

This results is autoboxing. You are converting int(primitive type) to it's corresponding wrapper.

 Integer i3 = new Integer(1); 

Here no need of autoboxing as you are directly creating an Integer object.

Now in

i1 == i2 i1 == i3 

i2 and i3 are automatically unboxed and regular int comparison takes place which is why you get true.

Now consider

i2 == i3 

Here both i2 and i3 are Integer objects that you are comparing. Since both are different object(since you have used new operator) it will obviously give false. Note == operator checks if two references point to same object or not. Infact .equals() method if not overridden does the same thing.

It is same as saying

    Integer i2 = new Integer(1);     Integer i3 = new Integer(1);     System.out.println("i2 == i3 "+(i2==i3)); 

which will again give you false.