Related
This question's answers are a community effort. Edit existing answers to improve this post. It is not currently accepting new answers or interactions.
What issues / pitfalls must be considered when overriding equals and hashCode?
The theory (for the language lawyers and the mathematically inclined):
equals() (javadoc) must define an equivalence relation (it must be reflexive, symmetric, and transitive). In addition, it must be consistent (if the objects are not modified, then it must keep returning the same value). Furthermore, o.equals(null) must always return false.
hashCode() (javadoc) must also be consistent (if the object is not modified in terms of equals(), it must keep returning the same value).
The relation between the two methods is:
Whenever a.equals(b), then a.hashCode() must be same as b.hashCode().
In practice:
If you override one, then you should override the other.
Use the same set of fields that you use to compute equals() to compute hashCode().
Use the excellent helper classes EqualsBuilder and HashCodeBuilder from the Apache Commons Lang library. An example:
public class Person {
private String name;
private int age;
// ...
#Override
public int hashCode() {
return new HashCodeBuilder(17, 31). // two randomly chosen prime numbers
// if deriving: appendSuper(super.hashCode()).
append(name).
append(age).
toHashCode();
}
#Override
public boolean equals(Object obj) {
if (!(obj instanceof Person))
return false;
if (obj == this)
return true;
Person rhs = (Person) obj;
return new EqualsBuilder().
// if deriving: appendSuper(super.equals(obj)).
append(name, rhs.name).
append(age, rhs.age).
isEquals();
}
}
Also remember:
When using a hash-based Collection or Map such as HashSet, LinkedHashSet, HashMap, Hashtable, or WeakHashMap, make sure that the hashCode() of the key objects that you put into the collection never changes while the object is in the collection. The bulletproof way to ensure this is to make your keys immutable, which has also other benefits.
There are some issues worth noticing if you're dealing with classes that are persisted using an Object-Relationship Mapper (ORM) like Hibernate, if you didn't think this was unreasonably complicated already!
Lazy loaded objects are subclasses
If your objects are persisted using an ORM, in many cases you will be dealing with dynamic proxies to avoid loading object too early from the data store. These proxies are implemented as subclasses of your own class. This means thatthis.getClass() == o.getClass() will return false. For example:
Person saved = new Person("John Doe");
Long key = dao.save(saved);
dao.flush();
Person retrieved = dao.retrieve(key);
saved.getClass().equals(retrieved.getClass()); // Will return false if Person is loaded lazy
If you're dealing with an ORM, using o instanceof Person is the only thing that will behave correctly.
Lazy loaded objects have null-fields
ORMs usually use the getters to force loading of lazy loaded objects. This means that person.name will be null if person is lazy loaded, even if person.getName() forces loading and returns "John Doe". In my experience, this crops up more often in hashCode() and equals().
If you're dealing with an ORM, make sure to always use getters, and never field references in hashCode() and equals().
Saving an object will change its state
Persistent objects often use a id field to hold the key of the object. This field will be automatically updated when an object is first saved. Don't use an id field in hashCode(). But you can use it in equals().
A pattern I often use is
if (this.getId() == null) {
return this == other;
}
else {
return this.getId().equals(other.getId());
}
But: you cannot include getId() in hashCode(). If you do, when an object is persisted, its hashCode changes. If the object is in a HashSet, you'll "never" find it again.
In my Person example, I probably would use getName() for hashCode and getId() plus getName() (just for paranoia) for equals(). It's okay if there are some risk of "collisions" for hashCode(), but never okay for equals().
hashCode() should use the non-changing subset of properties from equals()
A clarification about the obj.getClass() != getClass().
This statement is the result of equals() being inheritance unfriendly. The JLS (Java language specification) specifies that if A.equals(B) == true then B.equals(A) must also return true. If you omit that statement inheriting classes that override equals() (and change its behavior) will break this specification.
Consider the following example of what happens when the statement is omitted:
class A {
int field1;
A(int field1) {
this.field1 = field1;
}
public boolean equals(Object other) {
return (other != null && other instanceof A && ((A) other).field1 == field1);
}
}
class B extends A {
int field2;
B(int field1, int field2) {
super(field1);
this.field2 = field2;
}
public boolean equals(Object other) {
return (other != null && other instanceof B && ((B)other).field2 == field2 && super.equals(other));
}
}
Doing new A(1).equals(new A(1)) Also, new B(1,1).equals(new B(1,1)) result give out true, as it should.
This looks all very good, but look what happens if we try to use both classes:
A a = new A(1);
B b = new B(1,1);
a.equals(b) == true;
b.equals(a) == false;
Obviously, this is wrong.
If you want to ensure the symmetric condition. a=b if b=a and the Liskov substitution principle call super.equals(other) not only in the case of B instance, but check after for A instance:
if (other instanceof B )
return (other != null && ((B)other).field2 == field2 && super.equals(other));
if (other instanceof A) return super.equals(other);
else return false;
Which will output:
a.equals(b) == true;
b.equals(a) == true;
Where, if a is not a reference of B, then it might be a be a reference of class A (because you extend it), in this case you call super.equals() too.
For an inheritance-friendly implementation, check out Tal Cohen's solution, How Do I Correctly Implement the equals() Method?
Summary:
In his book Effective Java Programming Language Guide (Addison-Wesley, 2001), Joshua Bloch claims that "There is simply no way to extend an instantiable class and add an aspect while preserving the equals contract." Tal disagrees.
His solution is to implement equals() by calling another nonsymmetric blindlyEquals() both ways. blindlyEquals() is overridden by subclasses, equals() is inherited, and never overridden.
Example:
class Point {
private int x;
private int y;
protected boolean blindlyEquals(Object o) {
if (!(o instanceof Point))
return false;
Point p = (Point)o;
return (p.x == this.x && p.y == this.y);
}
public boolean equals(Object o) {
return (this.blindlyEquals(o) && o.blindlyEquals(this));
}
}
class ColorPoint extends Point {
private Color c;
protected boolean blindlyEquals(Object o) {
if (!(o instanceof ColorPoint))
return false;
ColorPoint cp = (ColorPoint)o;
return (super.blindlyEquals(cp) &&
cp.color == this.color);
}
}
Note that equals() must work across inheritance hierarchies if the Liskov Substitution Principle is to be satisfied.
Still amazed that none recommended the guava library for this.
//Sample taken from a current working project of mine just to illustrate the idea
#Override
public int hashCode(){
return Objects.hashCode(this.getDate(), this.datePattern);
}
#Override
public boolean equals(Object obj){
if ( ! obj instanceof DateAndPattern ) {
return false;
}
return Objects.equal(((DateAndPattern)obj).getDate(), this.getDate())
&& Objects.equal(((DateAndPattern)obj).getDate(), this.getDatePattern());
}
There are two methods in super class as java.lang.Object. We need to override them to custom object.
public boolean equals(Object obj)
public int hashCode()
Equal objects must produce the same hash code as long as they are equal, however unequal objects need not produce distinct hash codes.
public class Test
{
private int num;
private String data;
public boolean equals(Object obj)
{
if(this == obj)
return true;
if((obj == null) || (obj.getClass() != this.getClass()))
return false;
// object must be Test at this point
Test test = (Test)obj;
return num == test.num &&
(data == test.data || (data != null && data.equals(test.data)));
}
public int hashCode()
{
int hash = 7;
hash = 31 * hash + num;
hash = 31 * hash + (null == data ? 0 : data.hashCode());
return hash;
}
// other methods
}
If you want get more, please check this link as http://www.javaranch.com/journal/2002/10/equalhash.html
This is another example,
http://java67.blogspot.com/2013/04/example-of-overriding-equals-hashcode-compareTo-java-method.html
Have Fun! #.#
There are a couple of ways to do your check for class equality before checking member equality, and I think both are useful in the right circumstances.
Use the instanceof operator.
Use this.getClass().equals(that.getClass()).
I use #1 in a final equals implementation, or when implementing an interface that prescribes an algorithm for equals (like the java.util collection interfaces—the right way to check with with (obj instanceof Set) or whatever interface you're implementing). It's generally a bad choice when equals can be overridden because that breaks the symmetry property.
Option #2 allows the class to be safely extended without overriding equals or breaking symmetry.
If your class is also Comparable, the equals and compareTo methods should be consistent too. Here's a template for the equals method in a Comparable class:
final class MyClass implements Comparable<MyClass>
{
…
#Override
public boolean equals(Object obj)
{
/* If compareTo and equals aren't final, we should check with getClass instead. */
if (!(obj instanceof MyClass))
return false;
return compareTo((MyClass) obj) == 0;
}
}
For equals, look into Secrets of Equals by Angelika Langer. I love it very much. She's also a great FAQ about Generics in Java. View her other articles here (scroll down to "Core Java"), where she also goes on with Part-2 and "mixed type comparison". Have fun reading them!
equals() method is used to determine the equality of two objects.
as int value of 10 is always equal to 10. But this equals() method is about equality of two objects. When we say object, it will have properties. To decide about equality those properties are considered. It is not necessary that all properties must be taken into account to determine the equality and with respect to the class definition and context it can be decided. Then the equals() method can be overridden.
we should always override hashCode() method whenever we override equals() method. If not, what will happen? If we use hashtables in our application, it will not behave as expected. As the hashCode is used in determining the equality of values stored, it will not return the right corresponding value for a key.
Default implementation given is hashCode() method in Object class uses the internal address of the object and converts it into integer and returns it.
public class Tiger {
private String color;
private String stripePattern;
private int height;
#Override
public boolean equals(Object object) {
boolean result = false;
if (object == null || object.getClass() != getClass()) {
result = false;
} else {
Tiger tiger = (Tiger) object;
if (this.color == tiger.getColor()
&& this.stripePattern == tiger.getStripePattern()) {
result = true;
}
}
return result;
}
// just omitted null checks
#Override
public int hashCode() {
int hash = 3;
hash = 7 * hash + this.color.hashCode();
hash = 7 * hash + this.stripePattern.hashCode();
return hash;
}
public static void main(String args[]) {
Tiger bengalTiger1 = new Tiger("Yellow", "Dense", 3);
Tiger bengalTiger2 = new Tiger("Yellow", "Dense", 2);
Tiger siberianTiger = new Tiger("White", "Sparse", 4);
System.out.println("bengalTiger1 and bengalTiger2: "
+ bengalTiger1.equals(bengalTiger2));
System.out.println("bengalTiger1 and siberianTiger: "
+ bengalTiger1.equals(siberianTiger));
System.out.println("bengalTiger1 hashCode: " + bengalTiger1.hashCode());
System.out.println("bengalTiger2 hashCode: " + bengalTiger2.hashCode());
System.out.println("siberianTiger hashCode: "
+ siberianTiger.hashCode());
}
public String getColor() {
return color;
}
public String getStripePattern() {
return stripePattern;
}
public Tiger(String color, String stripePattern, int height) {
this.color = color;
this.stripePattern = stripePattern;
this.height = height;
}
}
Example Code Output:
bengalTiger1 and bengalTiger2: true
bengalTiger1 and siberianTiger: false
bengalTiger1 hashCode: 1398212510
bengalTiger2 hashCode: 1398212510
siberianTiger hashCode: –1227465966
Logically we have:
a.getClass().equals(b.getClass()) && a.equals(b) ⇒ a.hashCode() == b.hashCode()
But not vice-versa!
One gotcha I have found is where two objects contain references to each other (one example being a parent/child relationship with a convenience method on the parent to get all children).
These sorts of things are fairly common when doing Hibernate mappings for example.
If you include both ends of the relationship in your hashCode or equals tests it's possible to get into a recursive loop which ends in a StackOverflowException.
The simplest solution is to not include the getChildren collection in the methods.
This question's answers are a community effort. Edit existing answers to improve this post. It is not currently accepting new answers or interactions.
What issues / pitfalls must be considered when overriding equals and hashCode?
The theory (for the language lawyers and the mathematically inclined):
equals() (javadoc) must define an equivalence relation (it must be reflexive, symmetric, and transitive). In addition, it must be consistent (if the objects are not modified, then it must keep returning the same value). Furthermore, o.equals(null) must always return false.
hashCode() (javadoc) must also be consistent (if the object is not modified in terms of equals(), it must keep returning the same value).
The relation between the two methods is:
Whenever a.equals(b), then a.hashCode() must be same as b.hashCode().
In practice:
If you override one, then you should override the other.
Use the same set of fields that you use to compute equals() to compute hashCode().
Use the excellent helper classes EqualsBuilder and HashCodeBuilder from the Apache Commons Lang library. An example:
public class Person {
private String name;
private int age;
// ...
#Override
public int hashCode() {
return new HashCodeBuilder(17, 31). // two randomly chosen prime numbers
// if deriving: appendSuper(super.hashCode()).
append(name).
append(age).
toHashCode();
}
#Override
public boolean equals(Object obj) {
if (!(obj instanceof Person))
return false;
if (obj == this)
return true;
Person rhs = (Person) obj;
return new EqualsBuilder().
// if deriving: appendSuper(super.equals(obj)).
append(name, rhs.name).
append(age, rhs.age).
isEquals();
}
}
Also remember:
When using a hash-based Collection or Map such as HashSet, LinkedHashSet, HashMap, Hashtable, or WeakHashMap, make sure that the hashCode() of the key objects that you put into the collection never changes while the object is in the collection. The bulletproof way to ensure this is to make your keys immutable, which has also other benefits.
There are some issues worth noticing if you're dealing with classes that are persisted using an Object-Relationship Mapper (ORM) like Hibernate, if you didn't think this was unreasonably complicated already!
Lazy loaded objects are subclasses
If your objects are persisted using an ORM, in many cases you will be dealing with dynamic proxies to avoid loading object too early from the data store. These proxies are implemented as subclasses of your own class. This means thatthis.getClass() == o.getClass() will return false. For example:
Person saved = new Person("John Doe");
Long key = dao.save(saved);
dao.flush();
Person retrieved = dao.retrieve(key);
saved.getClass().equals(retrieved.getClass()); // Will return false if Person is loaded lazy
If you're dealing with an ORM, using o instanceof Person is the only thing that will behave correctly.
Lazy loaded objects have null-fields
ORMs usually use the getters to force loading of lazy loaded objects. This means that person.name will be null if person is lazy loaded, even if person.getName() forces loading and returns "John Doe". In my experience, this crops up more often in hashCode() and equals().
If you're dealing with an ORM, make sure to always use getters, and never field references in hashCode() and equals().
Saving an object will change its state
Persistent objects often use a id field to hold the key of the object. This field will be automatically updated when an object is first saved. Don't use an id field in hashCode(). But you can use it in equals().
A pattern I often use is
if (this.getId() == null) {
return this == other;
}
else {
return this.getId().equals(other.getId());
}
But: you cannot include getId() in hashCode(). If you do, when an object is persisted, its hashCode changes. If the object is in a HashSet, you'll "never" find it again.
In my Person example, I probably would use getName() for hashCode and getId() plus getName() (just for paranoia) for equals(). It's okay if there are some risk of "collisions" for hashCode(), but never okay for equals().
hashCode() should use the non-changing subset of properties from equals()
A clarification about the obj.getClass() != getClass().
This statement is the result of equals() being inheritance unfriendly. The JLS (Java language specification) specifies that if A.equals(B) == true then B.equals(A) must also return true. If you omit that statement inheriting classes that override equals() (and change its behavior) will break this specification.
Consider the following example of what happens when the statement is omitted:
class A {
int field1;
A(int field1) {
this.field1 = field1;
}
public boolean equals(Object other) {
return (other != null && other instanceof A && ((A) other).field1 == field1);
}
}
class B extends A {
int field2;
B(int field1, int field2) {
super(field1);
this.field2 = field2;
}
public boolean equals(Object other) {
return (other != null && other instanceof B && ((B)other).field2 == field2 && super.equals(other));
}
}
Doing new A(1).equals(new A(1)) Also, new B(1,1).equals(new B(1,1)) result give out true, as it should.
This looks all very good, but look what happens if we try to use both classes:
A a = new A(1);
B b = new B(1,1);
a.equals(b) == true;
b.equals(a) == false;
Obviously, this is wrong.
If you want to ensure the symmetric condition. a=b if b=a and the Liskov substitution principle call super.equals(other) not only in the case of B instance, but check after for A instance:
if (other instanceof B )
return (other != null && ((B)other).field2 == field2 && super.equals(other));
if (other instanceof A) return super.equals(other);
else return false;
Which will output:
a.equals(b) == true;
b.equals(a) == true;
Where, if a is not a reference of B, then it might be a be a reference of class A (because you extend it), in this case you call super.equals() too.
For an inheritance-friendly implementation, check out Tal Cohen's solution, How Do I Correctly Implement the equals() Method?
Summary:
In his book Effective Java Programming Language Guide (Addison-Wesley, 2001), Joshua Bloch claims that "There is simply no way to extend an instantiable class and add an aspect while preserving the equals contract." Tal disagrees.
His solution is to implement equals() by calling another nonsymmetric blindlyEquals() both ways. blindlyEquals() is overridden by subclasses, equals() is inherited, and never overridden.
Example:
class Point {
private int x;
private int y;
protected boolean blindlyEquals(Object o) {
if (!(o instanceof Point))
return false;
Point p = (Point)o;
return (p.x == this.x && p.y == this.y);
}
public boolean equals(Object o) {
return (this.blindlyEquals(o) && o.blindlyEquals(this));
}
}
class ColorPoint extends Point {
private Color c;
protected boolean blindlyEquals(Object o) {
if (!(o instanceof ColorPoint))
return false;
ColorPoint cp = (ColorPoint)o;
return (super.blindlyEquals(cp) &&
cp.color == this.color);
}
}
Note that equals() must work across inheritance hierarchies if the Liskov Substitution Principle is to be satisfied.
Still amazed that none recommended the guava library for this.
//Sample taken from a current working project of mine just to illustrate the idea
#Override
public int hashCode(){
return Objects.hashCode(this.getDate(), this.datePattern);
}
#Override
public boolean equals(Object obj){
if ( ! obj instanceof DateAndPattern ) {
return false;
}
return Objects.equal(((DateAndPattern)obj).getDate(), this.getDate())
&& Objects.equal(((DateAndPattern)obj).getDate(), this.getDatePattern());
}
There are two methods in super class as java.lang.Object. We need to override them to custom object.
public boolean equals(Object obj)
public int hashCode()
Equal objects must produce the same hash code as long as they are equal, however unequal objects need not produce distinct hash codes.
public class Test
{
private int num;
private String data;
public boolean equals(Object obj)
{
if(this == obj)
return true;
if((obj == null) || (obj.getClass() != this.getClass()))
return false;
// object must be Test at this point
Test test = (Test)obj;
return num == test.num &&
(data == test.data || (data != null && data.equals(test.data)));
}
public int hashCode()
{
int hash = 7;
hash = 31 * hash + num;
hash = 31 * hash + (null == data ? 0 : data.hashCode());
return hash;
}
// other methods
}
If you want get more, please check this link as http://www.javaranch.com/journal/2002/10/equalhash.html
This is another example,
http://java67.blogspot.com/2013/04/example-of-overriding-equals-hashcode-compareTo-java-method.html
Have Fun! #.#
There are a couple of ways to do your check for class equality before checking member equality, and I think both are useful in the right circumstances.
Use the instanceof operator.
Use this.getClass().equals(that.getClass()).
I use #1 in a final equals implementation, or when implementing an interface that prescribes an algorithm for equals (like the java.util collection interfaces—the right way to check with with (obj instanceof Set) or whatever interface you're implementing). It's generally a bad choice when equals can be overridden because that breaks the symmetry property.
Option #2 allows the class to be safely extended without overriding equals or breaking symmetry.
If your class is also Comparable, the equals and compareTo methods should be consistent too. Here's a template for the equals method in a Comparable class:
final class MyClass implements Comparable<MyClass>
{
…
#Override
public boolean equals(Object obj)
{
/* If compareTo and equals aren't final, we should check with getClass instead. */
if (!(obj instanceof MyClass))
return false;
return compareTo((MyClass) obj) == 0;
}
}
For equals, look into Secrets of Equals by Angelika Langer. I love it very much. She's also a great FAQ about Generics in Java. View her other articles here (scroll down to "Core Java"), where she also goes on with Part-2 and "mixed type comparison". Have fun reading them!
equals() method is used to determine the equality of two objects.
as int value of 10 is always equal to 10. But this equals() method is about equality of two objects. When we say object, it will have properties. To decide about equality those properties are considered. It is not necessary that all properties must be taken into account to determine the equality and with respect to the class definition and context it can be decided. Then the equals() method can be overridden.
we should always override hashCode() method whenever we override equals() method. If not, what will happen? If we use hashtables in our application, it will not behave as expected. As the hashCode is used in determining the equality of values stored, it will not return the right corresponding value for a key.
Default implementation given is hashCode() method in Object class uses the internal address of the object and converts it into integer and returns it.
public class Tiger {
private String color;
private String stripePattern;
private int height;
#Override
public boolean equals(Object object) {
boolean result = false;
if (object == null || object.getClass() != getClass()) {
result = false;
} else {
Tiger tiger = (Tiger) object;
if (this.color == tiger.getColor()
&& this.stripePattern == tiger.getStripePattern()) {
result = true;
}
}
return result;
}
// just omitted null checks
#Override
public int hashCode() {
int hash = 3;
hash = 7 * hash + this.color.hashCode();
hash = 7 * hash + this.stripePattern.hashCode();
return hash;
}
public static void main(String args[]) {
Tiger bengalTiger1 = new Tiger("Yellow", "Dense", 3);
Tiger bengalTiger2 = new Tiger("Yellow", "Dense", 2);
Tiger siberianTiger = new Tiger("White", "Sparse", 4);
System.out.println("bengalTiger1 and bengalTiger2: "
+ bengalTiger1.equals(bengalTiger2));
System.out.println("bengalTiger1 and siberianTiger: "
+ bengalTiger1.equals(siberianTiger));
System.out.println("bengalTiger1 hashCode: " + bengalTiger1.hashCode());
System.out.println("bengalTiger2 hashCode: " + bengalTiger2.hashCode());
System.out.println("siberianTiger hashCode: "
+ siberianTiger.hashCode());
}
public String getColor() {
return color;
}
public String getStripePattern() {
return stripePattern;
}
public Tiger(String color, String stripePattern, int height) {
this.color = color;
this.stripePattern = stripePattern;
this.height = height;
}
}
Example Code Output:
bengalTiger1 and bengalTiger2: true
bengalTiger1 and siberianTiger: false
bengalTiger1 hashCode: 1398212510
bengalTiger2 hashCode: 1398212510
siberianTiger hashCode: –1227465966
Logically we have:
a.getClass().equals(b.getClass()) && a.equals(b) ⇒ a.hashCode() == b.hashCode()
But not vice-versa!
One gotcha I have found is where two objects contain references to each other (one example being a parent/child relationship with a convenience method on the parent to get all children).
These sorts of things are fairly common when doing Hibernate mappings for example.
If you include both ends of the relationship in your hashCode or equals tests it's possible to get into a recursive loop which ends in a StackOverflowException.
The simplest solution is to not include the getChildren collection in the methods.
package MavenWeb.MavenWeb;
import java.util.HashMap;
import java.util.Map;
public class StringEquality {
public static void main(String[] args) {
Person p1 = new Person("Naveen", 22, 1000);
Person p2 = new Person("Naveen", 21, 2000);
Person p3 = new Person("Naveen", 23, 3000);
if(p1.equals(p2)){
System.out.println("P1 and p2 :" + p1.equals(p2));
} else{
System.out.println("P1 and p2 :" + p1.equals(p2));
}
if(p1.equals(p3)){
System.out.println("P1 and p3 :" + p1.equals(p3));
}
Map<Person, Object> map = new HashMap<Person, Object>();
map.put(p1, p1);
map.put(p2, p2);
System.out.println(map.get(new Person("Naveen", 21, 2000)));
}
}
...
class Person{
public Person(String name, int id, float salary){
this.name = name;
this.id = id;
this.salary = salary;
}
public Float getSalary() {
return salary;
}
public void setSalary(Float salary) {
this.salary = salary;
}
String name;
Float salary;
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public int getId() {
return id;
}
public void setId(int id) {
this.id = id;
}
int id;
#Override
public boolean equals(Object obj) {
if(obj == null){
return false;
}
if(this == obj){
return true;
}
if(obj instanceof Person){
Person person = (Person)obj;
if((person.getName().equals(name)) && person.getId() == id
&& person.getSalary() == salary){
return true;
}
}
return false;
}
#Override
public int hashCode() {
int hashCode = 1;
hashCode = 31 * hashCode + name.hashCode();
hashCode = 31 * hashCode + id;
hashCode = 31 * hashCode + salary.intValue();
return hashCode;
}
#Override
public String toString() {
return "Name :" + name + ", id : " + id;
}
}
I have write the following program. Little bit confused regarding hashcode implementation,
Why we use 31 in implementation
According to equals method logic this program should return p2(Naveen, 21) instance but it is returning null. why?
You're getting null from the map because the salary is stored as a Float (and not float), and you're comparing Float instances with == instead of comparing them with equals().
So the equals() method of Person checks that exactly the same Float instance is in both Person instances, instead of checking that the value of their salary is equal. You should use float (or better: double), unless the salary is nullable (you should also consider using BigDecimal to deal with exact money amounts). But in that case, you'll have to check for null in the Person.equals() and use equals() to compare the salaries. The easiest way, in Java 7 or later, is to use Objects.equals() to compare nullable objects together:
#Override
public boolean equals(Object obj) {
if(obj == null){
return false;
}
if(this == obj){
return true;
}
if(obj instanceof Person){
Person person = (Person)obj;
return Objects.equals(name, person.name)
&& id == person.id
&& Objects.equals(salary, person.salary);
}
return false;
}
#Override
public int hashCode() {
return Objects.hash(name, id, salary);
}
You used the number 31 so it is up to you what makes sense.
The number 31 is used for String as it is an odd prime (which produces more random hash codes) and larger than the number of letters in the alphabet so different letters are less likely to produce the same hash code for different text. With 31 you can have all 5 characters Strings being unique if they use either A-Z or a-z.
However, for longer strings I find that 57 works better for me and for non Strings other, larger prime numbers like 10191 could be better. I like 10191 as 101, 1019 and 10191 are prime numbers.
The problem is that your equals implementation is false.
Consider this :
Person p4 = new Person("Naveen", 21, 2000);
System.out.println(p2.equals(p4));
It prints false, or it should print true.
Or map.get(Object key) need the equals method to check if the key is already store in the map.
More formally, if this map contains a mapping from a key k to a value
v such that (key==null ? k==null : key.equals(k))
What's your problem with your equals method ?
In your equals method, you should replace
person.getSalary() == salary
By
1. person.getSalary().equals(salary)
2. person.getSalary().floatValue() == salary.floatValue()
Because getSalary() returns a Float object, so == will check their references and not their values.
There is no need of a the Float wrapper class, you should use float salary; or double salary; (if you want more precision).
You can use the equals implementations generated by Eclipse :
#Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Person other = (Person) obj;
if (id != other.id)
return false;
if (name == null) {
if (other.name != null)
return false;
} else if (!name.equals(other.name))
return false;
if (salary == null) {
if (other.salary != null)
return false;
} else if (!salary.equals(other.salary))
return false;
return true;
}
System.out.println(map.get(new Person("Naveen", 21, 2000)));
prints
Name :Naveen, id : 21
you can read this:
http://www.xyzws.com/javafaq/why-always-override-hashcode-if-overriding-equals/20
Why always override hashcode() if overriding equals()?
In Java, every object has access to the equals() method because it is inherited from the Object class. However, this default implementation just simply compares the memory addresses of the objects. You can override the default implementation of the equals() method defined in java.lang.Object. If you override the equals(), you MUST also override hashCode(). Otherwise a violation of the general contract for Object.hashCode will occur, which can have unexpected repercussions when your class is in conjunction with all hash-based collections.
Here is the contract, copied from the java.lang.Object specialization:
public int hashCode()
Returns a hash code value for the object. This method is supported for the benefit of hashtables such as those provided by java.util.Hashtable.
The general contract of hashCode is:
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 hashtables.
As much as is reasonably practical, the hashCode method defined by class Object does return distinct integers for distinct objects. (This is typically implemented by converting the internal address of the object into an integer, but this implementation technique is not required by the JavaTM programming language.)
The default implementation of equals() method checks to see if the two objects have the same identity. Similarly, the default implementation of the hashCode() method returns an integer based on the object's identity and is not based on the values of instance (and class) variables of the object. No matter how many times the values of its instance variables (data fields) change, the hash code calculated by the default hashCode implementation does not change during the life of the object.
Consider the following code, we have overridden equals() method to check if two objects are equal based on the values of their instance variables. Two objects may be stored at different memory addresses but may still be equal base on their instance variable.
public class CustomerID {
private long crmID;
private int nameSpace;
public CustomerID(long crmID, int nameSpace) {
super();
this.crmID = crmID;
this.nameSpace = nameSpace;
}
public boolean equals(Object obj) {
//null instanceof Object will always return false
if (!(obj instanceof CustomerID))
return false;
if (obj == this)
return true;
return this.crmID == ((CustomerID) obj).crmID &&
this.nameSpace == ((CustomerID) obj).nameSpace;
}
public static void main(String[] args) {
Map m = new HashMap();
m.put(new CustomerID(2345891234L,0),"Jeff Smith");
System.out.println(m.get(new CustomerID(2345891234L,0)));
}
}
Compile and run the above code, the output result is
null
What is wrong? The two instances of CustomerID are logically equal according to the class's equals method. Because the hashCode() method is not overridden, these two instances' identities are not in common to the default hashCode implementation. Therefore, the Object.hashCode returns two seemingly random numbers instead of two equal numbers. Such behavior violates "Equal objects must have equal hash codes" rule defined in the hashCode contract.
Let's provide a simple hashCode() method to fix this problem:
public class CustomerID {
private long crmID;
private int nameSpace;
public CustomerID(long crmID, int nameSpace) {
super();
this.crmID = crmID;
this.nameSpace = nameSpace;
}
public boolean equals(Object obj) {
//null instanceof Object will always return false
if (!(obj instanceof CustomerID))
return false;
if (obj == this)
return true;
return this.crmID == ((CustomerID) obj).crmID &&
this.nameSpace == ((CustomerID) obj).nameSpace;
}
public int hashCode() {
int result = 0;
result = (int)(crmID/12) + nameSpace;
return result;
}
public static void main(String[] args) {
Map m = new HashMap();
m.put(new CustomerID(2345891234L,0),"Jeff Smith");
System.out.println(m.get(new CustomerID(2345891234L,0)));
}
}
Compile and run the above code, the output result is
Jeff Smith
The hashcode distribution for instances of a class should be random. This is exactly what is meant by the third provision of the hashCode contract. Write a correct hashCode method is easy, but to write an effective hashCode method is extremely difficult.
For example, From How to Avoid Traps and Correctly Override Methods From java.lang.Object: If you are unsure how to implement hashCode(), just always return 0 in your implementations. So all of your custom objects will return the same hash code. Yes, it turns hashtable of your objects into one (possibly) long linked-list, but you have implemented hashCode() correctly!
public int hashCode(){
return 0;
}
It's legal because it ensures that equal objects have the same hash code, but it also indicates that every object has the same hash code. So every object will be hashed into the same bucket, and hash tables degenerate to linked lists. The performance is getting worse when it needs to process a large number of objects. How to implement a good hash function is a big topic and we will not cover here
I am trying to override equals method in Java. I have a class People which basically has 2 data fields name and age. Now I want to override equals method so that I can check between 2 People objects.
My code is as follows
public boolean equals(People other){
boolean result;
if((other == null) || (getClass() != other.getClass())){
result = false;
} // end if
else{
People otherPeople = (People)other;
result = name.equals(other.name) && age.equals(other.age);
} // end else
return result;
} // end equals
But when I write age.equals(other.age) it gives me error as equals method can only compare String and age is Integer.
Solution
I used == operator as suggested and my problem is solved.
//Written by K#stackoverflow
public class Main {
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
ArrayList<Person> people = new ArrayList<Person>();
people.add(new Person("Subash Adhikari", 28));
people.add(new Person("K", 28));
people.add(new Person("StackOverflow", 4));
people.add(new Person("Subash Adhikari", 28));
for (int i = 0; i < people.size() - 1; i++) {
for (int y = i + 1; y <= people.size() - 1; y++) {
boolean check = people.get(i).equals(people.get(y));
System.out.println("-- " + people.get(i).getName() + " - VS - " + people.get(y).getName());
System.out.println(check);
}
}
}
}
//written by K#stackoverflow
public class Person {
private String name;
private int age;
public Person(String name, int age){
this.name = name;
this.age = age;
}
#Override
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (obj.getClass() != this.getClass()) {
return false;
}
final Person other = (Person) obj;
if ((this.name == null) ? (other.name != null) : !this.name.equals(other.name)) {
return false;
}
if (this.age != other.age) {
return false;
}
return true;
}
#Override
public int hashCode() {
int hash = 3;
hash = 53 * hash + (this.name != null ? this.name.hashCode() : 0);
hash = 53 * hash + this.age;
return hash;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}
Output:
run:
-- Subash Adhikari - VS - K false
-- Subash Adhikari - VS - StackOverflow false
-- Subash Adhikari - VS - Subash Adhikari true
-- K - VS - StackOverflow false
-- K - VS - Subash Adhikari false
-- StackOverflow - VS - Subash Adhikari false
-- BUILD SUCCESSFUL (total time: 0 seconds)
Introducing a new method signature that changes the parameter types is called overloading:
public boolean equals(People other){
Here People is different than Object.
When a method signature remains the identical to that of its superclass, it is called overriding and the #Override annotation helps distinguish the two at compile-time:
#Override
public boolean equals(Object other){
Without seeing the actual declaration of age, it is difficult to say why the error appears.
I'm not sure of the details as you haven't posted the whole code, but:
remember to override hashCode() as well
the equals method should have Object, not People as its argument type. At the moment you are overloading, not overriding, the equals method, which probably isn't what you want, especially given that you check its type later.
you can use instanceof to check it is a People object e.g. if (!(other instanceof People)) { result = false;}
equals is used for all objects, but not primitives. I think you mean age is an int (primitive), in which case just use ==. Note that an Integer (with a capital 'I') is an Object which should be compared with equals.
See What issues should be considered when overriding equals and hashCode in Java? for more details.
Item 10: Obey the general contract when overriding equals
According to Effective Java, Overriding the equals method seems simple, but there are many ways to get it wrong, and consequences can be dire. The easiest way to avoid problems is not to override the equals method, in which case each instance of the class is equal only to itself. This is the right thing to do if any of the following conditions apply:
Each instance of the class is inherently unique. This is true for classes such as Thread that represent active entities rather than values. The equals implementation provided by Object has exactly the right behavior for these classes.
There is no need for the class to provide a “logical equality” test. For example, java.util.regex.Pattern could have overridden equals to check whether two Pattern instances represented exactly the same regular expression, but the designers didn’t think that clients would need or want this functionality. Under these circumstances, the equals implementation inherited from Object is ideal.
A superclass has already overridden equals, and the superclass behavior is appropriate for this class. For example, most Set implementations inherit their equals implementation from AbstractSet, List implementations from AbstractList, and Map implementations from AbstractMap.
The class is private or package-private, and you are certain that its equals method will never be invoked. If you are extremely risk-averse, you can override the equals method to ensure that it isn’t invoked accidentally:
The equals method implements an equivalence relation. It has these properties:
Reflexive: For any non-null reference value x, x.equals(x) must return true.
Symmetric: For any non-null reference values x and y, x.equals(y) must return true if and only if y.equals(x) returns true.
Transitive: For any non-null reference values x, y, z, if x.equals(y) returns true and y.equals(z) returns true, then x.equals(z) must return true.
Consistent: For any non-null reference values x and y, multiple invocations of x.equals(y) must consistently return true or consistently return false, provided no information used in equals comparisons is modified.
For any non-null reference value x, x.equals(null) must return false.
Here’s a recipe for a high-quality equals method:
Use the == operator to check if the argument is a reference to this object. If so, return true. This is just a performance optimization but one that is worth doing if the comparison is potentially expensive.
Use the instanceof operator to check if the argument has the correct type. If not, return false. Typically, the correct type is the class in which the method occurs. Occasionally, it is some interface implemented by this class. Use an interface if the class implements an interface that refines the equals contract to permit comparisons across classes that implement the interface. Collection interfaces such as Set, List, Map, and Map.Entry have this property.
Cast the argument to the correct type. Because this cast was preceded by an instanceof test, it is guaranteed to succeed.
For each “significant” field in the class, check if that field of the argument matches the corresponding field of this object. If all these tests succeed, return true; otherwise, return false. If the type in Step 2 is an interface, you must access the argument’s fields via interface methods; if the type is a class, you may be able to access the fields directly, depending on their accessibility.
For primitive fields whose type is not float or double, use the == operator for comparisons; for object reference fields, call the equals method recursively; for float fields, use the static Float.compare(float, float) method; and for double fields, use Double.compare(double, double). The special treatment of float and double fields is made necessary by the existence of Float.NaN, -0.0f and the analogous double values; While you could compare float and double fields with the static methods Float.equals and Double.equals, this would entail autoboxing on every comparison, which would have poor performance. For array fields, apply these guidelines to each element. If every element in an array field is significant, use one of the Arrays.equals methods.
Some object reference fields may legitimately contain null. To avoid the possibility of a NullPointerException, check such fields for equality using the static method Objects.equals(Object, Object).
// Class with a typical equals method
public final class PhoneNumber {
private final short areaCode, prefix, lineNum;
public PhoneNumber(int areaCode, int prefix, int lineNum) {
this.areaCode = rangeCheck(areaCode, 999, "area code");
this.prefix = rangeCheck(prefix, 999, "prefix");
this.lineNum = rangeCheck(lineNum, 9999, "line num");
}
private static short rangeCheck(int val, int max, String arg) {
if (val < 0 || val > max)
throw new IllegalArgumentException(arg + ": " + val);
return (short) val;
}
#Override public boolean equals(Object o) {
if (o == this)
return true;
if (!(o instanceof PhoneNumber))
return false;
PhoneNumber pn = (PhoneNumber)o;
return pn.lineNum == lineNum && pn.prefix == prefix
&& pn.areaCode == areaCode;
}
... // Remainder omitted
}
#Override
public boolean equals(Object that){
if(this == that) return true;//if both of them points the same address in memory
if(!(that instanceof People)) return false; // if "that" is not a People or a childclass
People thatPeople = (People)that; // than we can cast it to People safely
return this.name.equals(thatPeople.name) && this.age == thatPeople.age;// if they have the same name and same age, then the 2 objects are equal unless they're pointing to different memory adresses
}
When comparing objects in Java, you make a semantic check, comparing the type and identifying state of the objects to:
itself (same instance)
itself (clone, or reconstructed copy)
other objects of different types
other objects of the same type
null
Rules:
Symmetry: a.equals(b) == b.equals(a)
equals() always yields true or false, but never a NullpointerException, ClassCastException or any other throwable
Comparison:
Type check: both instances need to be of the same type, meaning you have to compare the actual classes for equality. This is often not correctly implemented, when developers use instanceof for type comparison (which only works as long as there are no subclasses, and violates the symmetry rule when A extends B -> a instanceof b != b instanceof a).
Semantic check of identifying state: Make sure you understand by which state the instances are identified. Persons may be identified by their social security number, but not by hair color (can be dyed), name (can be changed) or age (changes all the time). Only with value objects should you compare the full state (all non-transient fields), otherwise check only what identifies the instance.
For your Person class:
public boolean equals(Object obj) {
// same instance
if (obj == this) {
return true;
}
// null
if (obj == null) {
return false;
}
// type
if (!getClass().equals(obj.getClass())) {
return false;
}
// cast and compare state
Person other = (Person) obj;
return Objects.equals(name, other.name) && Objects.equals(age, other.age);
}
Reusable, generic utility class:
public final class Equals {
private Equals() {
// private constructor, no instances allowed
}
/**
* Convenience equals implementation, does the object equality, null and type checking, and comparison of the identifying state
*
* #param instance object instance (where the equals() is implemented)
* #param other other instance to compare to
* #param stateAccessors stateAccessors for state to compare, optional
* #param <T> instance type
* #return true when equals, false otherwise
*/
public static <T> boolean as(T instance, Object other, Function<? super T, Object>... stateAccessors) {
if (instance == null) {
return other == null;
}
if (instance == other) {
return true;
}
if (other == null) {
return false;
}
if (!instance.getClass().equals(other.getClass())) {
return false;
}
if (stateAccessors == null) {
return true;
}
return Stream.of(stateAccessors).allMatch(s -> Objects.equals(s.apply(instance), s.apply((T) other)));
}
}
For your Person class, using this utility class:
public boolean equals(Object obj) {
return Equals.as(this, obj, t -> t.name, t -> t.age);
}
Since I'm guessing age is of type int:
public boolean equals(Object other){
boolean result;
if((other == null) || (getClass() != other.getClass())){
result = false;
} // end if
else{
People otherPeople = (People)other;
result = name.equals(otherPeople.name) && age == otherPeople.age;
} // end else
return result;
} // end equals
if age is int you should use == if it is Integer object then you can use equals().
You also need to implement hashcode method if you override equals. Details of the contract is available in the javadoc of Object and also at various pages in web.
tl;dr
record Person ( String name , int age ) {}
if(
new Person( "Carol" , 27 ) // Compiler auto-generates implicitly the constructor.
.equals( // Compiler auto-generates implicitly the `equals` method.
new Person( "Carol" , 42 )
)
) // Returns `false`, as the name matches but the age differs.
{ … }
Details
While your specific problem is solved (using == for equality test between int primitive values), there is an alternative that eliminates the need to write that code.
record
Java 16 brings the record feature.
A record is a brief way to write a class whose main purpose is to transparently and immutably carry data. The compiler implicitly creates the constructor, getters, equals & hashCode, and toString.
equals method provided automatically
The default implicit equals method compares each and every member field that you declared for the record. The members can be objects or primitives, both types are automatically compared in the default equals method.
For example, if you have a Person record carrying two fields, name & age, both of those fields are automatically compared to determine equality between a pair of Person objects.
public record Person ( String name , int age ) {}
Try it.
Person alice = new Person( "Alice" , 23 ) ;
Person alice2 = new Person( "Alice" , 23 ) ;
Person bob = new Person( "Bob" , 19 ) ;
boolean samePerson1 = alice.equals( alice2 ) ; // true.
boolean samePerson2 = alice.equals( bob ) ; // false.
You can override the equals method on a record, if you want a behavior other than the default. But if you do override equals, be sure to override hashCode for consistent logic, as you would for a conventional Java class. And, think twice: Whenever adding methods to a record, reconsider if a record structure is really appropriate to that problem domain.
Tip: A record can be defined within another class, and even locally within a method.
Here is the solution that I recently used:
public class Test {
public String a;
public long b;
public Date c;
public String d;
#Override
public boolean equals(Object obj) {
if (this == obj) {
return true;
}
if (!(obj instanceof Test)) {
return false;
}
Test testOther = (Test) obj;
return (a != null ? a.equals(testOther.a) : testOther.a == null)
&& (b == testOther.b)
&& (c != null ? c.equals(testOther.c) : testOther.c == null)
&& (d != null ? d.equals(testOther.d) : testOther.d == null);
}
}
For lazy programmers: lombok library is very easy and time saving. please have a look at this link
instead of writing lines of codes and rules, you just need to apply this library in your IDE and then just #Data and it is Done.
import lombok.Data;
#Data // this is the magic word :D
public class pojo {
int price;
String currency;
String productName;
}
in fact in the above code, #Data is a shortcut for
import lombok.Data;
import lombok.EqualsAndHashCode;
import lombok.Getter;
import lombok.Setter;
import lombok.ToString;
#Getter
#Setter
#EqualsAndHashCode
#ToString
//or instead of all above #Data
public class pojo {
int price;
String currency;
String productName;
}
This question's answers are a community effort. Edit existing answers to improve this post. It is not currently accepting new answers or interactions.
What issues / pitfalls must be considered when overriding equals and hashCode?
The theory (for the language lawyers and the mathematically inclined):
equals() (javadoc) must define an equivalence relation (it must be reflexive, symmetric, and transitive). In addition, it must be consistent (if the objects are not modified, then it must keep returning the same value). Furthermore, o.equals(null) must always return false.
hashCode() (javadoc) must also be consistent (if the object is not modified in terms of equals(), it must keep returning the same value).
The relation between the two methods is:
Whenever a.equals(b), then a.hashCode() must be same as b.hashCode().
In practice:
If you override one, then you should override the other.
Use the same set of fields that you use to compute equals() to compute hashCode().
Use the excellent helper classes EqualsBuilder and HashCodeBuilder from the Apache Commons Lang library. An example:
public class Person {
private String name;
private int age;
// ...
#Override
public int hashCode() {
return new HashCodeBuilder(17, 31). // two randomly chosen prime numbers
// if deriving: appendSuper(super.hashCode()).
append(name).
append(age).
toHashCode();
}
#Override
public boolean equals(Object obj) {
if (!(obj instanceof Person))
return false;
if (obj == this)
return true;
Person rhs = (Person) obj;
return new EqualsBuilder().
// if deriving: appendSuper(super.equals(obj)).
append(name, rhs.name).
append(age, rhs.age).
isEquals();
}
}
Also remember:
When using a hash-based Collection or Map such as HashSet, LinkedHashSet, HashMap, Hashtable, or WeakHashMap, make sure that the hashCode() of the key objects that you put into the collection never changes while the object is in the collection. The bulletproof way to ensure this is to make your keys immutable, which has also other benefits.
There are some issues worth noticing if you're dealing with classes that are persisted using an Object-Relationship Mapper (ORM) like Hibernate, if you didn't think this was unreasonably complicated already!
Lazy loaded objects are subclasses
If your objects are persisted using an ORM, in many cases you will be dealing with dynamic proxies to avoid loading object too early from the data store. These proxies are implemented as subclasses of your own class. This means thatthis.getClass() == o.getClass() will return false. For example:
Person saved = new Person("John Doe");
Long key = dao.save(saved);
dao.flush();
Person retrieved = dao.retrieve(key);
saved.getClass().equals(retrieved.getClass()); // Will return false if Person is loaded lazy
If you're dealing with an ORM, using o instanceof Person is the only thing that will behave correctly.
Lazy loaded objects have null-fields
ORMs usually use the getters to force loading of lazy loaded objects. This means that person.name will be null if person is lazy loaded, even if person.getName() forces loading and returns "John Doe". In my experience, this crops up more often in hashCode() and equals().
If you're dealing with an ORM, make sure to always use getters, and never field references in hashCode() and equals().
Saving an object will change its state
Persistent objects often use a id field to hold the key of the object. This field will be automatically updated when an object is first saved. Don't use an id field in hashCode(). But you can use it in equals().
A pattern I often use is
if (this.getId() == null) {
return this == other;
}
else {
return this.getId().equals(other.getId());
}
But: you cannot include getId() in hashCode(). If you do, when an object is persisted, its hashCode changes. If the object is in a HashSet, you'll "never" find it again.
In my Person example, I probably would use getName() for hashCode and getId() plus getName() (just for paranoia) for equals(). It's okay if there are some risk of "collisions" for hashCode(), but never okay for equals().
hashCode() should use the non-changing subset of properties from equals()
A clarification about the obj.getClass() != getClass().
This statement is the result of equals() being inheritance unfriendly. The JLS (Java language specification) specifies that if A.equals(B) == true then B.equals(A) must also return true. If you omit that statement inheriting classes that override equals() (and change its behavior) will break this specification.
Consider the following example of what happens when the statement is omitted:
class A {
int field1;
A(int field1) {
this.field1 = field1;
}
public boolean equals(Object other) {
return (other != null && other instanceof A && ((A) other).field1 == field1);
}
}
class B extends A {
int field2;
B(int field1, int field2) {
super(field1);
this.field2 = field2;
}
public boolean equals(Object other) {
return (other != null && other instanceof B && ((B)other).field2 == field2 && super.equals(other));
}
}
Doing new A(1).equals(new A(1)) Also, new B(1,1).equals(new B(1,1)) result give out true, as it should.
This looks all very good, but look what happens if we try to use both classes:
A a = new A(1);
B b = new B(1,1);
a.equals(b) == true;
b.equals(a) == false;
Obviously, this is wrong.
If you want to ensure the symmetric condition. a=b if b=a and the Liskov substitution principle call super.equals(other) not only in the case of B instance, but check after for A instance:
if (other instanceof B )
return (other != null && ((B)other).field2 == field2 && super.equals(other));
if (other instanceof A) return super.equals(other);
else return false;
Which will output:
a.equals(b) == true;
b.equals(a) == true;
Where, if a is not a reference of B, then it might be a be a reference of class A (because you extend it), in this case you call super.equals() too.
For an inheritance-friendly implementation, check out Tal Cohen's solution, How Do I Correctly Implement the equals() Method?
Summary:
In his book Effective Java Programming Language Guide (Addison-Wesley, 2001), Joshua Bloch claims that "There is simply no way to extend an instantiable class and add an aspect while preserving the equals contract." Tal disagrees.
His solution is to implement equals() by calling another nonsymmetric blindlyEquals() both ways. blindlyEquals() is overridden by subclasses, equals() is inherited, and never overridden.
Example:
class Point {
private int x;
private int y;
protected boolean blindlyEquals(Object o) {
if (!(o instanceof Point))
return false;
Point p = (Point)o;
return (p.x == this.x && p.y == this.y);
}
public boolean equals(Object o) {
return (this.blindlyEquals(o) && o.blindlyEquals(this));
}
}
class ColorPoint extends Point {
private Color c;
protected boolean blindlyEquals(Object o) {
if (!(o instanceof ColorPoint))
return false;
ColorPoint cp = (ColorPoint)o;
return (super.blindlyEquals(cp) &&
cp.color == this.color);
}
}
Note that equals() must work across inheritance hierarchies if the Liskov Substitution Principle is to be satisfied.
Still amazed that none recommended the guava library for this.
//Sample taken from a current working project of mine just to illustrate the idea
#Override
public int hashCode(){
return Objects.hashCode(this.getDate(), this.datePattern);
}
#Override
public boolean equals(Object obj){
if ( ! obj instanceof DateAndPattern ) {
return false;
}
return Objects.equal(((DateAndPattern)obj).getDate(), this.getDate())
&& Objects.equal(((DateAndPattern)obj).getDate(), this.getDatePattern());
}
There are two methods in super class as java.lang.Object. We need to override them to custom object.
public boolean equals(Object obj)
public int hashCode()
Equal objects must produce the same hash code as long as they are equal, however unequal objects need not produce distinct hash codes.
public class Test
{
private int num;
private String data;
public boolean equals(Object obj)
{
if(this == obj)
return true;
if((obj == null) || (obj.getClass() != this.getClass()))
return false;
// object must be Test at this point
Test test = (Test)obj;
return num == test.num &&
(data == test.data || (data != null && data.equals(test.data)));
}
public int hashCode()
{
int hash = 7;
hash = 31 * hash + num;
hash = 31 * hash + (null == data ? 0 : data.hashCode());
return hash;
}
// other methods
}
If you want get more, please check this link as http://www.javaranch.com/journal/2002/10/equalhash.html
This is another example,
http://java67.blogspot.com/2013/04/example-of-overriding-equals-hashcode-compareTo-java-method.html
Have Fun! #.#
There are a couple of ways to do your check for class equality before checking member equality, and I think both are useful in the right circumstances.
Use the instanceof operator.
Use this.getClass().equals(that.getClass()).
I use #1 in a final equals implementation, or when implementing an interface that prescribes an algorithm for equals (like the java.util collection interfaces—the right way to check with with (obj instanceof Set) or whatever interface you're implementing). It's generally a bad choice when equals can be overridden because that breaks the symmetry property.
Option #2 allows the class to be safely extended without overriding equals or breaking symmetry.
If your class is also Comparable, the equals and compareTo methods should be consistent too. Here's a template for the equals method in a Comparable class:
final class MyClass implements Comparable<MyClass>
{
…
#Override
public boolean equals(Object obj)
{
/* If compareTo and equals aren't final, we should check with getClass instead. */
if (!(obj instanceof MyClass))
return false;
return compareTo((MyClass) obj) == 0;
}
}
For equals, look into Secrets of Equals by Angelika Langer. I love it very much. She's also a great FAQ about Generics in Java. View her other articles here (scroll down to "Core Java"), where she also goes on with Part-2 and "mixed type comparison". Have fun reading them!
equals() method is used to determine the equality of two objects.
as int value of 10 is always equal to 10. But this equals() method is about equality of two objects. When we say object, it will have properties. To decide about equality those properties are considered. It is not necessary that all properties must be taken into account to determine the equality and with respect to the class definition and context it can be decided. Then the equals() method can be overridden.
we should always override hashCode() method whenever we override equals() method. If not, what will happen? If we use hashtables in our application, it will not behave as expected. As the hashCode is used in determining the equality of values stored, it will not return the right corresponding value for a key.
Default implementation given is hashCode() method in Object class uses the internal address of the object and converts it into integer and returns it.
public class Tiger {
private String color;
private String stripePattern;
private int height;
#Override
public boolean equals(Object object) {
boolean result = false;
if (object == null || object.getClass() != getClass()) {
result = false;
} else {
Tiger tiger = (Tiger) object;
if (this.color == tiger.getColor()
&& this.stripePattern == tiger.getStripePattern()) {
result = true;
}
}
return result;
}
// just omitted null checks
#Override
public int hashCode() {
int hash = 3;
hash = 7 * hash + this.color.hashCode();
hash = 7 * hash + this.stripePattern.hashCode();
return hash;
}
public static void main(String args[]) {
Tiger bengalTiger1 = new Tiger("Yellow", "Dense", 3);
Tiger bengalTiger2 = new Tiger("Yellow", "Dense", 2);
Tiger siberianTiger = new Tiger("White", "Sparse", 4);
System.out.println("bengalTiger1 and bengalTiger2: "
+ bengalTiger1.equals(bengalTiger2));
System.out.println("bengalTiger1 and siberianTiger: "
+ bengalTiger1.equals(siberianTiger));
System.out.println("bengalTiger1 hashCode: " + bengalTiger1.hashCode());
System.out.println("bengalTiger2 hashCode: " + bengalTiger2.hashCode());
System.out.println("siberianTiger hashCode: "
+ siberianTiger.hashCode());
}
public String getColor() {
return color;
}
public String getStripePattern() {
return stripePattern;
}
public Tiger(String color, String stripePattern, int height) {
this.color = color;
this.stripePattern = stripePattern;
this.height = height;
}
}
Example Code Output:
bengalTiger1 and bengalTiger2: true
bengalTiger1 and siberianTiger: false
bengalTiger1 hashCode: 1398212510
bengalTiger2 hashCode: 1398212510
siberianTiger hashCode: –1227465966
Logically we have:
a.getClass().equals(b.getClass()) && a.equals(b) ⇒ a.hashCode() == b.hashCode()
But not vice-versa!
One gotcha I have found is where two objects contain references to each other (one example being a parent/child relationship with a convenience method on the parent to get all children).
These sorts of things are fairly common when doing Hibernate mappings for example.
If you include both ends of the relationship in your hashCode or equals tests it's possible to get into a recursive loop which ends in a StackOverflowException.
The simplest solution is to not include the getChildren collection in the methods.