HashSet Duplicate element ?

[Soumya Ranjan Mohanty]

   import java.util.*;

public class WrapStr {
   
   private String s;
   public WrapStr(String s) {
   this.s=s;
   }
public String getS() {
  return s;
  }
  
  public static void main(String args[]) {
   
   HashSet<Object> hs=new HashSet<Object>();
   
   WrapStr ws1=new WrapStr("john");
   WrapStr ws2=new WrapStr("john");
   
   String s1=new String("john");
   String s2=new String("john");
   
   hs.add(ws1);
   hs.add(ws2);
   hs.add(s1);
   hs.add(s2);
   
   System.out.println(hs.size());
   }
 }
  

How to modify the above code such that the hs.add(ws2); returns false. i.e ws1 and ws2 will be considered as equal(because both have the same value of s) and cannot be added into HashSet .

[Christophe Verré]

You have to override both Object#equals and Object#hashCode. Can you figure out why and how ?

[Soumya Ranjan Mohanty]

import java.util.*;

public class WrapStr {
   
   private String s;
   public WrapStr(String s) {
   this.s=s;
   }
public String getS() {
  return s;
  }
  
  public boolean equals(Object o) {
  
    if ((o instanceof WrapStr) && this.getS()==(((WrapStr)o).getS()))
      {
        return true;
      }
    else { 
   	   return false; 
   	  }
    }
  
  public int hashcode(){
  	return 5;
  }
  
   
   public static void main(String args[]) {
   
   HashSet<Object> hs=new HashSet<Object>();
   
   WrapStr ws1=new WrapStr("john");
   WrapStr ws2=new WrapStr("john");
   
   String s1=new String("john");
   String s2=new String("john");
   
  System.out.println(hs.add(ws1));
  System.out.println( hs.add(ws2));
  System.out.println(hs.add(s1));
  System.out.println(hs.add(s2));
   
   System.out.println(hs.size());
   }
 }
  

I am getting the same output.

[Christophe Verré]

Are you sure that comparing String values with "==" is right ?

[Christophe Verré]

Also, you are not overriding hashCode. Try to use the @Override annotation on your method to see what happens.

[Elchin Asgarli]

Christophe Verré wrote:Are you sure that comparing String values with "==" is right ?

Theoretically it could, since he is hardcoding "john" and JVM must cache it, so == could work.

[Elchin Asgarli]

Even thought == on strings could theoretically work, you better use .equals on them, so instead of doing this.getS()==(((WrapStr)o).getS()) do this.getS().equals((((WrapStr)o).getS())). This approach is safer and not JVM-dependent

[Abimaran Kugathasan]

Check this code:

import java.util.*;
import java.text.*;
public class test6 {

	String count;
	test6(String s)	{
		count = s;
	}

	public static void main(String[] args) {
	HashSet<test6> set=new HashSet<test6>();
	System.out.println(set.add(new test6("5")));
	System.out.println(set.add(new test6("5")));
	System.out.println(set.add(new test6("3")));


	}
	public boolean equals(Object o)	{
		return ((this.count).equals( ((test6)o).count));
	}
	public int hashCode()	{
		return 434;
	}
}

You will get them....

[Rob Spoor]

Why not return count.hashCode() from your hashCode method? A fixed value is terrible for performance when used with a HashMap or HashSet.

[Abimaran Kugathasan]

Rob Prime wrote:Why not return count.hashCode() from your hashCode method? A fixed value is terrible for performance when used with a HashMap or HashSet.

I just follow her coding, in that she uses a fixed value.

[Rob Spoor]

Ah right. Then my comment applies to her code

[Sandra Bachan]

I modified WrapStr program such that the equals() method uses equals() instead of == and it also outputs size as 3 instead of 2. Code is below, please guide.

import java.util.*;

public class WrapStr{
	private String s;
	public WrapStr(String s){
		this.s = s;
	}
	public String getS(){
		return s;
	}
	
	public boolean equals(Object o){
		if (o instanceof WrapStr)
			return (this.getS()).equals(((WrapStr)o).getS());
		else
			return false;
	}
	public int hashcode(){
		return 5;
	}
	
	public static void main(String[] args){
		HashSet<Object> hs = new HashSet<Object>();
		
		
		WrapStr ws1 = new WrapStr("john");
		WrapStr ws2 = new WrapStr("john");
		
		String s1 = new String("john");
		String s2 = new String("john");
		
		hs.add(ws1);
		hs.add(ws2);
		hs.add(s1);
		hs.add(s2);
		
		System.out.println(hs.size());
	}
}

[Rob Spoor]

I'll quote Christophe just because he's 100% right:

Christophe Verré wrote:Also, you are not overriding hashCode. Try to use the @Override annotation on your method to see what happens.

[Abimaran Kugathasan]

Rob Prime wrote:I'll quote Christophe just because he's 100% right:

Christophe Verré wrote:Also, you are not overriding hashCode. Try to use the @Override annotation on your method to see what happens.

We use these annotations in EJB, What is the need of it here? It's just another overriding?

[Rob Spoor]

Just try it, you'll see how useful @Override can be.