+ 2

Can we override method(int) as method(Integer) like in the below example?

public class MainClass { public static void main(String[] args) { Integer i1 = 127; Integer i2 = 127; System.out.println(i1 == i2); Integer i3 = 128; Integer i4 = 128; System.out.println(i3 == i4); } }

28th Dec 2016, 8:12 PM
Ankit RAJ
Ankit RAJ - avatar
2 Respuestas
+ 3
Hi Ankit, int is a primitive and Integer is an object. If you instantiate an Integer object you should use this sintax: Integer x=new Integer(1); from java 1.5 there is a unboxing functionality that instantiate automatically the object: Integer x=1; Now the question: Integer a=1; Integer b=1; (a==b) -> false! Because a and b are two different objects! (a.equals(b)) -> true! Because if you should compare two objects you must use a specific implementation - "equals" that in this case would do a simple comparison == between the two int values. You could use this rule in more complex objects.. Try to create your own Integer class where implementing the equals method (for an hint take a look at java code).
28th Dec 2016, 9:10 PM
Michael Isac Girardi
Michael Isac Girardi - avatar
+ 1
No. It gives compile time error. Compiler treats int and Integer as two different types while overriding. Auto-boxing doesn’t happen here.
28th Dec 2016, 8:13 PM
Ankit RAJ
Ankit RAJ - avatar