LinkedHashSet

[swapna mallipudi]

The Question is from CertPal. What is the output of the program?

 publci class Lion{
String name="";
public static void main(String[] args){
Lion l=new Lion(); l.name="lion";
Lion l2=new Lion(); l2.name="Lion";
Lion l3=new Lion(); l3.name="lion";
String s1="lion";
String s2="Lion";
String s3="lion";
Set set=new LinkedHashSet();
set.add(l); set.add(l2); set.add(l3);
set.add(s1); set.add(s2); set.add(s3);
System.out.println(set.size());
}
public String toString(){
 return name;
}
}
   

Size is 5. I don't know what I am missing here. I was thinking LinkedHashSet doesn't allow duplicates. Then howcome the size is 5.

Thank you.

[Christophe Verré]

I was thinking LinkedHashSet doesn't allow duplicates.

How do you know if a Lion is equal to another ?

[Harpreet Singh janda]

In case of set it always first check the HashCode and the use the equals to compare the objects.
Lion class does not override the hashcode and equals method so two objects of lion class will never be equal until they are not pointing to same object. So there will not be any duplicate lion object but same is not true for String objects.

If you try the below code :

public class Lion{
	String name="";
	public static void main(String[] args){
		Lion l=new Lion(); l.name="lion";
		Lion l2=new Lion(); l2.name="Lion";
		Lion l3=new Lion(); l3.name="lion";
		
		String s1="lion";
		String s2="Lion";
		String s3="lion";
		Set set=new LinkedHashSet();
		set.add(l); set.add(l2); set.add(l3);
		set.add(s1); set.add(s2); set.add(s3);
		System.out.println(set.size());
		
		for( Iterator i = set.iterator(); i.hasNext();)	{
			Object ob = i.next();
			System.out.println(ob + " : " + ob.hashCode());
		}
	}
}

Output will be

   5
Lion@10b62c9 : 17523401
Lion@82ba41 : 8567361
Lion@923e30 : 9584176
lion : 3321884
Lion : 2368572 

See, there are 3 lion objects and 2 String objects Because String override equals method but Lion does not. If you override the equals and hashcode methods in Lion class you will get desired results.

[Harpreet Singh janda]

Here is the modified version of above code:

   public class Lion{
	String name="";
	public static void main(String[] args){
		Lion l=new Lion(); l.name="lion";
		Lion l2=new Lion(); l2.name="Lion";
		Lion l3=new Lion(); l3.name="lion";
		
		String s1="lion";
		String s2="Lion";
		String s3="lion";
		Set set=new LinkedHashSet();
		set.add(l); set.add(l2); set.add(l3);
		set.add(s1); set.add(s2); set.add(s3);
		
		System.out.println(set.size());
		
		for( Iterator i = set.iterator(); i.hasNext();)	{
			Object ob = i.next();
			System.out.println(ob + " : " + ob.hashCode());
		}
	}
	/*public String toString(){
		return name;
	}*/
	
	@Override
	public boolean equals(Object obj) {
		
		if ( obj instanceof Lion)
		return name.equals(((Lion)obj).name);
			
		return name.equals(((String)obj));
	}
	
	@Override
	public int hashCode() {
		return name.hashCode();
	}
} 

and the output is

    4
Lion@32b01c : 3321884
Lion@24243c : 2368572
lion : 3321884
Lion : 2368572

So the crux is : Whenever you are using an object with collections, always take care of hashcode and equals methods

[Shailesh Phatak]

Harpreet Singh janda wrote:So the crux is : Whenever you are using an object with collections, always take care of hashcode and equals methods

It was a great help Harpreet

[jose chiramal]

I am wondering how overriding equals() and hashcode() makes such a huge difference ?? Whats so special about equals and hashcode, that we should override them in user defined classes ??

Also which are those classes that override equals and hashcode , besides String and Wrapper Classes ?

[Rajeev Trikha]

For two objects of a given class to be equal, they must have the same hashcode and meaningful equivalence defined within equal method. The default implementation in Object only gives two objects to be equal if they are of the same type and share the common reference.

For the exam purposes, it is useful to know that StringBuffer doesn't define equal method.