Java hashCode() and equals() – Contract, rules and best practices

By Lokesh Gupta | Filed Under: Java Basics

Learn about Java hashCode() and equals() methods, their default implementation and how to correctly override them. Also learn to implement these methods using Apache Commons package’s utility classes HashCodeBuilder and EqualsBuilder.

hashCode() and equals() methods have been defined in Object class which is parent class for java objects. For this reason, all java objects inherit a default implementation of these methods.

Table of Contents:

1) Usage of hashCode() and equals() Methods
2) Override the default behavior
3) EqualsBuilder and HashCodeBuilder
4) Generate hashCode() and equals() using Eclipse
5) Important things to remember
6) Special Attention When Using in ORM

1. Usage of hashCode() and equals() Methods

equals(Object otherObject) – As method name suggests, is used to simply verify the equality of two objects. It’s default implementation simply check the object references of two objects to verify their equality. By default, two objects are equal if and only if they are stored in the same memory address.
hashcode() – Returns a unique integer value for the object in runtime. By default, integer value is mostly derived from memory address of the object in heap (but it’s not mandatory always).
This hash code is used for determining the bucket location, when this object needs to be stored in some HashTable like data structure.

1.1. Contract between hashCode() and equals()

It is generally necessary to override the hashCode() method whenever equals() method is overridden, so as to maintain the general contract for the hashCode() method, which states that equal objects must have equal hash codes.

Whenever it is invoked on the same object more than once during an execution of a Java application, the hashCode method must consistently return the same integer, provided no information used in equals comparisons on the object is modified.
This integer need not remain consistent from one execution of an application to another execution of the same application.
If two objects are equal according to the equals(Object) method, then calling the hashCode method on each of the two objects must produce the same integer result.
It is not required that if two objects are unequal according to the equals(java.lang.Object) method, then calling the hashCode method on each of the two objects must produce distinct integer results.
However, the programmer should be aware that producing distinct integer results for unequal objects may improve the performance of hash tables.

2. Override the default behavior of hashCode() and equals()

Everything works fine until you do not override any of these methods in your classes. But, sometimes the application needs to change the default behavior of some objects. Lets understand why we need to override equals and hashcode methods.

2.1. Default Behavior

Lets take an example where your application has Employee object. Lets create a minimal possible structure of Employee class:

public class Employee
{
	private Integer id;
	private String firstname;
	private String lastName;
	private String department;

	//Setters and Getters
}

Above Employee class has some very basic attributes and their accessor methods. Now consider a simple situation where you need to compare two employee objects.

public class EqualsTest {
	public static void main(String[] args) {
		Employee e1 = new Employee();
		Employee e2 = new Employee();

		e1.setId(100);
		e2.setId(100);

		System.out.println(e1.equals(e2));	//false
	}
}

No prize for guessing. Above method will print "false". But, is it really correct after knowing that both objects represent the same employee. In a real-time application, this should return true.

2.2. Should we override only equals() method?

To achieve correct application behavior, we need to override equals() method as below:

public boolean equals(Object o) {
	if(o == null)
	{
		return false;
	}
	if (o == this)
	{
		return true;
	}
	if (getClass() != o.getClass())
	{
		return false;
	}
	
	Employee e = (Employee) o;
	return (this.getId() == e.getId());
}

Add this method to your Employee class, and EqualsTest will start returning "true".

So are we done? Not yet. Lets test again above modified Employee class in different way.

import java.util.HashSet;
import java.util.Set;

public class EqualsTest
{
	public static void main(String[] args)
	{
		Employee e1 = new Employee();
		Employee e2 = new Employee();

		e1.setId(100);
		e2.setId(100);

		//Prints 'true'
		System.out.println(e1.equals(e2));

		Set<Employee> employees = new HashSet<Employee>();
		employees.add(e1);
		employees.add(e2);
		
		System.out.println(employees);	//Prints two objects
	}
}

Above class prints two objects in the second print statement. If both employee objects have been equal, in a Set which stores only unique objects, there must be only one instance inside HashSet, after all both objects refer to the same employee. What is it we are missing??

2.3. Override hashCode() method also

We are missing the second important method hashCode(). As java docs say, if you override equals() method then you must override hashCode() method. So lets add another method in our Employee class.

@Override
public int hashCode()
{
	final int PRIME = 31;
	int result = 1;
	result = PRIME * result + getId();
	return result;
}

Once above method is added in Employee class, the second statement start printing only single object in second statement, and thus validating the true equality of e1 and e2.

3. EqualsBuilder and HashCodeBuilder utility classes

Apache commons provide two excellent utility classes HashCodeBuilder and EqualsBuilder for generating hash code and equals methods. Below is its usage:

import org.apache.commons.lang3.builder.EqualsBuilder;
import org.apache.commons.lang3.builder.HashCodeBuilder;
public class Employee
{
	private Integer id;
	private String firstname;
	private String lastName;
	private String department;
	
	//Setters and Getters

	@Override
	public int hashCode()
	{
		final int PRIME = 31;
		return new HashCodeBuilder(getId()%2==0?getId()+1:getId(), PRIME).toHashCode();
	}

	@Override
	public boolean equals(Object o) {
	if (o == null)
	   return false;
	   
	if (o == this)
	   return true;
	   
	if (o.getClass() != getClass())
	   return false;
	
	Employee e = (Employee) o;
	
	return new EqualsBuilder().
			  append(getId(), e.getId()).
			  isEquals();
	}
}

4. Generate hashCode() and equals() using Eclipse

If you are using any code editor, then most editors also are capable of generating some good structure for you. For example, Eclipse IDE has option to generate a very good implementation of hashCode() and equals() for you.

Right click on java file -> Source -> Generate hashCode() and equals() …

5. Java hashCode() and equals() best practices

Always use same attributes of an object to generate hashCode() and equals() both. As in our case, we have used employee id.
equals() must be consistent (if the objects are not modified, then it must keep returning the same value).
Whenever a.equals(b), then a.hashCode() must be same as b.hashCode().
If you override one, then you should override the other.

6. Special Attention When Using in ORM

If you’re dealing with an ORM, make sure to always use getters, and never field references in hashCode() and equals(). This is for reason, in ORM, occasionally fields are lazy loaded and not available until called their getter methods.

For example, In our Employee class if we use e1.id == e2.id. It is very much possible that id field is lazy loaded. So in this case, one might be zero or null, and thus resulting in incorrect behavior.

But if uses e1.getId() == e2.getId(), we can be sure even if field is lazy loaded; calling getter will populate the field first.

This is all i know about hashCode() and equals() methods. I hope, it will help someone somewhere.

If you feel, I am missing something or wrong somewhere, please leave a comment. I will update this post again to help others.

Happy Learning !!

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About Lokesh Gupta

A family guy with fun loving nature. Love computers, programming and solving everyday problems. Find me on Facebook and Twitter.

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Gopi

Hi Lokesh,

I was confused when it comes to Strings, Let see the bellow code

package com.mytest.garbage;
public class StringTest {
	public static void main(String [] ar){		
		String str = "hello";
		String str1 = new String("hello");		
		if(str == str1){
			System.out.println("both are equal");
		}else{
			System.out.println("both are not  equal");
		}
		
		System.out.println("hashcode of str "+ str.hashCode());
		System.out.println("hashcode of str1 "+ str1.hashCode());		
	}	
}

Output:
————-
both are not equal
hashcode of str 99162322
hashcode of str1 99162322

Here hash code for both the string are same but they are not equal.Can you please help me out from this.

Thanks in advance!

Kuldeep Bisen

As you know that two different objects can have the same hashcode. this is a case and that’s the reason yo equal method here to compare the content. use == to compare primitive type, not wrapper type.

Bernard

For String Class .equals() method and hashCode() method are already overridden. So when you compare hashcode of 2 strings with same content as in your case you will get same hashcode. But when you compare strings with == it performs reference comparisons for str and str1 in your case. Since the references are different thats why == gives false. If you compare str and str1 with .equals() you will get result as true.

sachin rokade

why PRIME value is 31

sindhu

Hi, your blog really helps a lot for all, I appreciate for sharing your knowledge with us.
I am having a small problem, i would be glad if you help me out.

While i am working with hashmap<String,List>, my requirement is to read data from file and set to the bean class, then i need to get the data from bean class and based on conditions i need to add to the list then add to the map. In each if- block i am adding to the map–>map.put(stringvalue,listobject), in each if-block in map entry value is getting replaced with new value and again new key value is getting returned.
like–>when i added first record in first if block
1) v->first record
when again i add to map the second record in another second if block then
2) [v—>{first record,second record},M–>{second record}]
But my requirement is like
v->{first record} M–>{second record}
and when finally i retrieve from map all the values are getting replaced with the latest values.
please suggest me with a solution.

@anonymous

Very nice article. I understood clearly about the contract between equals and hashcode methods.

tushar

Thanks for the elaborative article

Lakshman

Another Important Note you can add is:

– If two objects are not equal according to

equals(Object)

then it doesn’t necessarily mean that their hashcode must be different.

Raj kiran

good one.somehow i understand

Marcos

when in doubt, use constant value for hashCode

vaibhav

Where did you find that hashCode method return memory address of an object?

1. Usage of hashCode() and equals() Methods

1.1. Contract between hashCode() and equals()

2. Override the default behavior of hashCode() and equals()

2.1. Default Behavior

2.2. Should we override only equals() method?

2.3. Override hashCode() method also

3. EqualsBuilder and HashCodeBuilder utility classes

4. Generate hashCode() and equals() using Eclipse

5. Java hashCode() and equals() best practices

6. Special Attention When Using in ORM

About Lokesh Gupta

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