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.
Let's say I create two objects in my Main, through the same constructor, containing the same type of 5 integers.
Secondclass mp1 = new Secondclass ( 0 , 1 , 2 , 3 , 4 );
Secondclass mp2 = new Secondclass ( 5 , 6 , 7 , 8 , 9 );
And in my Secondclass I have a method called Comparing , where I wanna pass the two objects and compare some of the integers , somewhat like this :
mp1.Comparing(mp1,mp2);
Is this possible in Java ?
I'm new to programming, I hope I was clear enough.
Jon Skeet has already given you an answer in his comment, but let me elaborate a little on what he has said with some code examples:
Convention has it that you would define a method int compareTo(SecondClass other) in SecondClass such that for o1.compareTo(o2) it returns
a negative value if o1 is considered smaller than o2
0 if they are equal
a positive value if o1 is considered greater than o2
For an easier example, let's assume SecondClass had only one int which would be the field value. Then it could be implemented as follows:
public int compareTo(SecondClass other) {
return this.value - other.value;
}
If you do this, you can (and should) make your SecondClass implement Comparable.
Also, you should make sure that o1.compareTo(o2) == 0 if and only if o1.equals(o2), so make sure you implement a correct equals method. In the example, it could look like this:
public boolean equals(Object other) {
(if other == null || !(other instanceof SecondClass))
return false;
else
return this.value == ((SecondClass) other).value;
}
(You could also use compareTo for this of course, by replacing the return statement with return compareTo((SecondClass) other) == 0.
And last but not least, if you have overriden the equals method, always make sure to override hashCode as well, such that if two instances of your class are equal, they have the same hash code. In the example, you could simply use the integer value:
public int hashCode() {
return value;
}
Complete code
public class SecondClass implements Comparable<SecondClass> {
private int value;
public int compareTo(SecondClass other) {
return this.value - other.value;
}
public boolean equals(Object other) {
(if other == null || !(other instanceof SecondClass))
return false;
else
return this.value == ((SecondClass) other).value;
}
public int hashCode() {
return value;
}
}
That's all you need to do.
You can implement Comparator
public SecondClassComparator implements Comparator<SecondClass> {
public int compare (SecondClass c1, SecondClass c2) {
// Do some comparision
}
}
Java has Comparator(http://docs.oracle.com/javase/7/docs/api/java/util/Comparator.html) interface for the same. You can implement comparator interface in SecondClass class.
object1.compare(object1, object2)
but as object1 data is already available, I will prefer implementing comparable (http://docs.oracle.com/javase/7/docs/api/java/lang/Comparable.html) interface.
You can just use it as you use equals method in java
object1.compareTo(object2)
I have a test class like so:
public class CompareObjects {
public static class Obj {
public int i;
public Obj(int i) {
this.i = i;
}
}
public static void main(String[] args) {
Obj o1 = new Obj(0);
Obj o2 = new Obj(0);
if(o1 == o2) {
System.out.println("Equal");
}else{
System.out.println("Not equal");
}
}
}
I though the test would return "Equal", but it didn't. Why doesn't Java consider two objects with equal components not the same? Am I doing something wrong here? I have a feeling I completely overlooked something when I started learning Java.
Also, I tested the two against each other with the equals() method, and I get the same result. My reason for this question is that I would like to be able to test an ArrayList with the contains() method to see if one object has the same components as another and therefore equal. Any ideas?
== compares the references to the object. For example:
Obj a = new Obj(0);
Obj b = a;
//a == b
Try implementing equals():
public static class Obj {
public int i;
public Obj(int i) {
this.i = i;
}
#Override
public boolean equals(Object other) {
if (other == this) return true;
if (!(other instanceof Obj) || other == null) return false;
return i == ((Obj)other).i;
}
#Override
public int hashCode() {
return i;
}
}
Then, you can use if(o1.equals(o2)) {. However, this is not really a good example, read this (link) for more information.
== returns true only if you are comparing the same object [i.e. the same memory location].
If you want to compare objects by their fields, you have to overload the equals() method, in order to induce an equivalence relation over them.
public boolean equals(Object other){
return this.i == other.i;
}
Be sure that the equals() method respects reflexivity, symmmetry, transitivity.
== compares the reference equality, i.e if they refer to the same object in the memory.
You need to override equals() method, and use it whenever you want to compare their values. Also, override hashCode() which is used by HashMap for example.
The == operator does not check for equality in class data; rather, it checks to see if they are the same location in memory. If you did o2 = o1 instead of initializing them the same way, they would be the same location in memory, so o2==o1 would return true. However, since they were initialized some separately, it returns false. Instead, you should define an equals method and implement that.
I'm having a problem with getting an ArrayList to correctly use an overriden equals. the problem is that I'm trying to use the equals to only test for a single key field, and using ArrayList.contains() to test for the existence of an object with the correct field. Here is an example
public class TestClass {
private static class InnerClass{
private final String testKey;
//data and such
InnerClass(String testKey, int dataStuff) {
this.testKey =testKey;
//etc
}
#Override
public boolean equals (Object in) {
System.out.println("reached here");
if(in == null) {
return false;
}else if( in instanceof String) {
String inString = (String) in;
return testKey == null ? false : testKey.equals(inString);
}else {
return false;
}
}
}
public static void main(String[] args) {
ArrayList<InnerClass> objectList = new ArrayList<InnerClass>();
//add some entries
objectList.add(new InnerClass("UNIQUE ID1", 42));
System.out.println( objectList.contains("UNIQUE ID1"));
}
}
What worries me is that not only am I getting false on the output, but I'm also not getting the "reached here" output.
Does anyone have any ideas why this override is being completely ignored? Is there some subtlety with overrides and inner classes I don't know of?
Edit:
Having problems with the site so I cant seem to mark the answered.
Thanks for the quick response: yes an oversight on my part that it is the String .equals thta is called, not my custom one. I guess it's old fashioned checks for now
If you check sources of ArrayList, you will see that it calls equals of other object. In your case it will call equals of String "UNIQUE ID1" which will check that other object is not of type String and just returns false:
public boolean contains(Object o) {
return indexOf(o) >= 0;
}
public int indexOf(Object o) {
...
for (int i = 0; i < size; i++)
if (o.equals(elementData[i]))
return i;
...
return -1;
}
For your case call contains with InnerClass that only contains id:
objectList.contains(new InnerClass("UNIQUE ID1"))
Don't forget to implement equals for InnerClass which compares id only.
According to the JavaDoc of List.contains(o), it is defined to return true
if and only if this list contains at least one element e such that (o==null ? e==null : o.equals(e)).
Note that this definition calls equals on o, which is the parameter and not the element that is in the List.
Therefore String.equals() will be called and not InnerClass.equals().
Also note that the contract for Object.equals() states that
It is symmetric: for any non-null reference values x and y, x.equals(y) should return true if and only if y.equals(x) returns true.
But you violate this constraint, since new TestClass("foo", 1).equals("foo") returns true but "foo".equals(new TestClass("foo", 1)) will always return false.
Unfortunately this means that your use case (a custom class that can be equal to another standard class) can not be implemented in a completely conforming way.
If you still want to do something like this, you'll have to read the specification (and sometimes the implementation) of all your collection classes very carefully and check for pitfalls such as this.
You're invoking contains with an argument that's a String and not an InnerClass:
System.out.println( objectList.contains("UNIQUE ID1"))
In my JDK:
public class ArrayList {
public boolean contains(Object o) {
return indexOf(o) >= 0;
}
public int indexOf(Object o) {
if (o == null) {
// omitted for brevity - aix
} else {
for (int i = 0; i < size; i++)
if (o.equals(elementData[i])) // <<<<<<<<<<<<<<<<<<<<<<
return i;
}
return -1;
}
}
Note how indexOf calls o.equals(). In your case, o is a String, so your objectList.contains will be using String.equals and not InnerClass.equals.
Generally, you need to also override hashCode() but this is not the main problem here. You are having an asymmetric equals(..) method. The docs make it clear that it should be symmetric:
It is symmetric: for any non-null reference values x and y, x.equals(y) should return true if and only if y.equals(x) returns true.
And what you observe is an unexpected behaviour due to broken contract.
Create an utility method that iterates all items and verifies with equals(..) on the string:
public static boolean containsString(List<InnerClass> items, String str) {
for (InnerClass item : items) {
if (item.getTestKey().equals(str)) {
return true;
}
}
return false;
}
You can do a similar thing with guava's Iterables.any(..) method:
final String str = "Foo";
boolean contains = Iterables.any(items, new Predicate<InnerClass>() {
#Override
public boolean apply(InnerClass input){
return input.getTestKey().equals(str);
}
}
Your equals implementation is wrong. Your in parameter should not be a String. It should be an InnerClass.
public boolean equals(Object o) {
if (this == o) return true;
if (!(o instanceof InnerClass) return false;
InnerClass that = (InnerClass)o;
// check for null keys if you need to
return this.testKey.equals(that.testKey);
}
(Note that instanceof null returns false, so you don't need to check for null first).
You would then test for existence of an equivalent object in your list using:
objectList.contains(new InnerClass("UNIQUE ID1"));
But if you really want to check for InnerClass by String key, why not use Map<String,InnerClass> instead?
Although not answering your question, many Collections use hashcode(). You should override that too to "agree" with equals().
Actually, you should always implement both equals and hashcode together, and they should always be consistent with each other. As the javadoc for Object.equals() states:
Note that it is generally necessary to
override the hashCode method whenever
this method is overridden, so as to
maintain the general contract for the
hashCode method, which states that
equal objects must have equal hash
codes.
Specifically, many Collections rely on this contract being upheld - behaviour is undefined otherwise.
There are a few issues with your code. My suggestion would be to avoid overriding the equals entirely if you are not familiar with it and extend it into a new implementation like so...
class MyCustomArrayList extends ArrayList<InnerClass>{
public boolean containsString(String value){
for(InnerClass item : this){
if (item.getString().equals(value){
return true;
}
}
return false;
}
}
Then you can do something like
List myList = new MyCustomArrayList()
myList.containsString("some string");
I suggest this because if you override the equals should also override the hashCode and it seems you are lacking a little knowledge in this area - so i would just avoid it.
Also, the contains method calls the equals method which is why you are seeing the "reached here". Again if you don't understand the call flow i would just avoid it.
in the other way, your equal method gets called if you change your code as follows. hope this clears the concept.
package com.test;
import java.util.ArrayList;
import java.util.List;
public class TestClass {
private static class InnerClass{
private final String testKey;
//data and such
InnerClass(String testKey, int dataStuff) {
this.testKey =testKey;
//etc
}
#Override
public boolean equals (Object in1) {
System.out.println("reached here");
if(in1 == null) {
return false;
}else if( in1 instanceof InnerClass) {
return ((InnerClass) this).testKey == null ? false : ((InnerClass) this).testKey.equals(((InnerClass) in1).testKey);
}else {
return false;
}
}
}
public static void main(String[] args) {
ArrayList<InnerClass> objectList = new ArrayList<InnerClass>();
InnerClass in1 = new InnerClass("UNIQUE ID1", 42);
InnerClass in2 = new InnerClass("UNIQUE ID1", 42);
//add some entries
objectList.add(in1);
System.out.println( objectList.contains(in2));
}
}
As many posts have said, the problem is that list.indexOf(obj) function calls "equals" of the obj, not the items on the list.
I had the same problem and "contains()" didn't satisfy me, as I need to know where is the element!. My aproach is to create an empty element with just the parameter to compare, and then call indexOf.
Implement a function like this,
public static InnerClass empty(String testKey) {
InnerClass in = new InnerClass();
in.testKey =testKey;
return in;
}
And then, call indexOf like this:
ind position = list.indexOf(InnerClass.empty(key));
There are two errors in your code.
First:
The "contains" method called on "objectList" object should pass a new InnerClass object as the parameter.
Second:
The equals method (should accept the parameter as Object, and is correct) should handle the code properly according to the received object.
Like this:
#Override
public boolean equals (Object in) {
System.out.println("reached here");
if(in == null) {
return false;
}else if( in instanceof InnerClass) {
String inString = ((InnerClass)in).testKey;
return testKey == null ? false : testKey.equals(inString);
}else {
return false;
}
}
This post was first written before Java 8 was available but now that it's 2017 instead of using the List.containts(...) method you can use the new Java 8 way like this:
System.out.println(objectList.stream().filter(obj -> obj.getTestKey().equals("UNIQUE ID1")).findAny().isPresent());
And give your TestClass a getter for your testKey field:
public String getTestKey() {
return testKey;
}
The benefit of this approach is that you don't have to modify the equals or hash method and you'll look like a boss to your peers!