Why is this HashSet allowing duplicates?

[Vidhya Ramaswamy]

import java.util.*;
 
public class Nearly {
String value;
Nearly(String v){ value = v;}
public int hashCode(){ return 1; }
public boolean equals(Nearly n){
if(value.charAt(0)==n.value.charAt(0))return true;
return false;
}
public static void main(String[] args) {
Nearly n1 = new Nearly("aaa");
Nearly n2 = new Nearly("aaa");
String s ="-";
if(n1.equals(n2))s+="1";
if(n1==n2)s+="2";
Set<Nearly> set = new HashSet<Nearly>();
set.add(n1);
set.add(n2);
System.out.println(s + " " + set.size());
}
}[B]Output: -1 2[/B]

Question taken from LearnKey MasterExam.
Why is the set allowing duplicates, and showing the size as 2?
I understand, it is because the equals() is not properly overridden, it doesn't take an Object. So, is the set allowing duplicate because of the equals() in Nearly class?

[Ankit Garg]

Yes! the equals method must take an Object parameter to make it a successful override and then it will not allow duplicates...

[Vidhya Ramaswamy]

Thanks Ankit for your quick response.
Is this what happens?

 if(n1.equals(n2))s+="1";   

calls equals() in the Nearly class.

  set.add(n1); 
set.add(n2);   

equals() and hashCode() are called before adding to HashSet. But here, the equals() called is of Object class(since Nearly class does not override equals() properly. So, n2 is added to the set.
Please correct me if I'm wrong.

[Ankit Garg]

You are right. But if you change your code to something like this, then see the output. Change this

if(n1.equals(n2))s+="1";

to this

Object obj = n2;
if(n1.equals(obj))s+="1";

Now you will see that the objects are now not equal as the equals method in Object class will be called now instead of equals in Nearly class...

[Preethi Dev]

Hi Ankit,
i tried this program ,as you have mentioned(i.e replacing Object b=n2)

but it still shows set.size() as 2. if it overrides equals() in object, then how it allows to add the second one?
please clear this too....

Thanks
Preetha

[Vidhya Ramaswamy]

Hi Preetha,

Replacing the code with
Object obj = n2;
if(n1.equals(obj))s+="1";

calls the Object class equals(), which just does a direct compare(==).
So, 1 is not printed in this case.

The reason it shows set.size() as 2 is because both the references do not refer to the same object, as(n1!=n2).

Try this out(assigns n1 to obj, so both refer to the same object):

Object obj = n1;
if(n1.equals(obj))s+="1"; //calls Object class equals()
if(n1==n2)s+="2";
Set<Nearly> set = new HashSet<Nearly>();
set.add(n1);
set.add((Nearly)obj); //trying to add duplicate
System.out.println(s + " " + set.size());

Now, set.size() will print 1.