Why do I need to override the equals and hashCode methods in Java?
Recently I read through this Developer Works Document.
The document is all about defining hashCode() and equals() effectively and correctly, however I am not able to figure out why we need to override these two methods.
How can I take the decision to implement these methods efficiently?
Solution 1:
Joshua Bloch says on Effective Java
You must override hashCode() in every class that overrides equals(). Failure to do so will result in a violation of the general contract for Object.hashCode(), which will prevent your class from functioning properly in conjunction with all hash-based collections, including HashMap, HashSet, and Hashtable.
Let's try to understand it with an example of what would happen if we override equals() without overriding hashCode() and attempt to use a Map.
Say we have a class like this and that two objects of MyClass are equal if their importantField is equal (with hashCode() and equals() generated by eclipse)
public class MyClass {
private final String importantField;
private final String anotherField;
public MyClass(final String equalField, final String anotherField) {
this.importantField = equalField;
this.anotherField = anotherField;
}
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result
+ ((importantField == null) ? 0 : importantField.hashCode());
return result;
}
@Override
public boolean equals(final Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
final MyClass other = (MyClass) obj;
if (importantField == null) {
if (other.importantField != null)
return false;
} else if (!importantField.equals(other.importantField))
return false;
return true;
}
}
Imagine you have this
MyClass first = new MyClass("a","first");
MyClass second = new MyClass("a","second");
Override only equals
If only equals is overriden, then when you call myMap.put(first,someValue) first will hash to some bucket and when you call myMap.put(second,someOtherValue) it will hash to some other bucket (as they have a different hashCode). So, although they are equal, as they don't hash to the same bucket, the map can't realize it and both of them stay in the map.
Although it is not necessary to override equals() if we override hashCode(), let's see what would happen in this particular case where we know that two objects of MyClass are equal if their importantField is equal but we do not override equals().
Override only hashCode
If you only override hashCode then when you call myMap.put(first,someValue) it takes first, calculates its hashCode and stores it in a given bucket. Then when you call myMap.put(second,someOtherValue) it should replace first with second as per the Map Documentation because they are equal (according to the business requirement).
But the problem is that equals was not redefined, so when the map hashes second and iterates through the bucket looking if there is an object k such that second.equals(k) is true it won't find any as second.equals(first) will be false.
Hope it was clear
Solution 2:
Collections such as HashMap and HashSet use a hashcode value of an object to determine how it should be stored inside a collection, and the hashcode is used again in order to locate the object
in its collection.
Hashing retrieval is a two-step process:
- Find the right bucket (using
hashCode()) - Search the bucket for the right element (using
equals())
Here is a small example on why we should overrride equals() and hashcode().
Consider an Employee class which has two fields: age and name.
public class Employee {
String name;
int age;
public Employee(String name, int age) {
this.name = name;
this.age = age;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
@Override
public boolean equals(Object obj) {
if (obj == this)
return true;
if (!(obj instanceof Employee))
return false;
Employee employee = (Employee) obj;
return employee.getAge() == this.getAge()
&& employee.getName() == this.getName();
}
// commented
/* @Override
public int hashCode() {
int result=17;
result=31*result+age;
result=31*result+(name!=null ? name.hashCode():0);
return result;
}
*/
}
Now create a class, insert Employee object into a HashSet and test whether that object is present or not.
public class ClientTest {
public static void main(String[] args) {
Employee employee = new Employee("rajeev", 24);
Employee employee1 = new Employee("rajeev", 25);
Employee employee2 = new Employee("rajeev", 24);
HashSet<Employee> employees = new HashSet<Employee>();
employees.add(employee);
System.out.println(employees.contains(employee2));
System.out.println("employee.hashCode(): " + employee.hashCode()
+ " employee2.hashCode():" + employee2.hashCode());
}
}
It will print the following:
false
employee.hashCode(): 321755204 employee2.hashCode():375890482
Now uncomment hashcode() method , execute the same and the output would be:
true
employee.hashCode(): -938387308 employee2.hashCode():-938387308
Now can you see why if two objects are considered equal, their hashcodes must
also be equal? Otherwise, you'd never be able to find the object since the default
hashcode method in class Object virtually always comes up with a unique number
for each object, even if the equals() method is overridden in such a way that two
or more objects are considered equal. It doesn't matter how equal the objects are if
their hashcodes don't reflect that. So one more time: If two objects are equal, their
hashcodes must be equal as well.
Solution 3:
You must override hashCode() in every class that overrides equals(). Failure to do so will result in a violation of the general contract for Object.hashCode(), which will prevent your class from functioning properly in conjunction with all hash-based collections, including HashMap, HashSet, and Hashtable.
from Effective Java, by Joshua Bloch
By defining equals() and hashCode() consistently, you can improve the usability of your classes as keys in hash-based collections. As the API doc for hashCode explains: "This method is supported for the benefit of hashtables such as those provided by java.util.Hashtable."
The best answer to your question about how to implement these methods efficiently is suggesting you to read Chapter 3 of Effective Java.
Solution 4:
Identity is not equality.
- equals operator
==test identity. -
equals(Object obj)method compares equality test(i.e. we need to tell equality by overriding the method)
Why do I need to override the equals and hashCode methods in Java?
First we have to understand the use of equals method.
In order to identity differences between two objects we need to override equals method.
For example:
Customer customer1=new Customer("peter");
Customer customer2=customer1;
customer1.equals(customer2); // returns true by JVM. i.e. both are refering same Object
------------------------------
Customer customer1=new Customer("peter");
Customer customer2=new Customer("peter");
customer1.equals(customer2); //return false by JVM i.e. we have two different peter customers.
------------------------------
Now I have overriden Customer class equals method as follows:
@Override
public boolean equals(Object obj) {
if (this == obj) // it checks references
return true;
if (obj == null) // checks null
return false;
if (getClass() != obj.getClass()) // both object are instances of same class or not
return false;
Customer other = (Customer) obj;
if (name == null) {
if (other.name != null)
return false;
} else if (!name.equals(other.name)) // it again using bulit in String object equals to identify the difference
return false;
return true;
}
Customer customer1=new Customer("peter");
Customer customer2=new Customer("peter");
Insteady identify the Object equality by JVM, we can do it by overring equals method.
customer1.equals(customer2); // returns true by our own logic
Now hashCode method can understand easily.
hashCode produces integer in order to store object in data structures like HashMap, HashSet.
Assume we have override equals method of Customer as above,
customer1.equals(customer2); // returns true by our own logic
While working with data structure when we store object in buckets(bucket is a fancy name for folder). If we use built-in hash technique, for above two customers it generates two different hashcode. So we are storing the same identical object in two different places. To avoid this kind of issues we should override the hashCode method also based on the following principles.
- un-equal instances may have same hashcode.
- equal instances should return same hashcode.
Solution 5:
Simply put, the equals-method in Object check for reference equality, where as two instances of your class could still be semantically equal when the properties are equal. This is for instance important when putting your objects into a container that utilizes equals and hashcode, like HashMap and Set. Let's say we have a class like:
public class Foo {
String id;
String whatevs;
Foo(String id, String whatevs) {
this.id = id;
this.whatevs = whatevs;
}
}
We create two instances with the same id:
Foo a = new Foo("id", "something");
Foo b = new Foo("id", "something else");
Without overriding equals we are getting:
- a.equals(b) is false because they are two different instances
- a.equals(a) is true since it's the same instance
- b.equals(b) is true since it's the same instance
Correct? Well maybe, if this is what you want. But let's say we want objects with the same id to be the same object, regardless if it's two different instances. We override the equals (and hashcode):
public class Foo {
String id;
String whatevs;
Foo(String id, String whatevs) {
this.id = id;
this.whatevs = whatevs;
}
@Override
public boolean equals(Object other) {
if (other instanceof Foo) {
return ((Foo)other).id.equals(this.id);
}
}
@Override
public int hashCode() {
return this.id.hashCode();
}
}
As for implementing equals and hashcode I can recommend using Guava's helper methods