Why do I need to override the equals and hashCode methods in Java?

Question

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?

Answer 1

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:

1. Find the right bucket (using hashCode())
2. 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.

Answer 2

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;     } } 

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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.

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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