After researching I still can't find the specific solution for my problem. I have an "approximately equals" method that uses an epsilon, while my hashCode method uses the exact values. This breaks the precondition of HashSet when I compare the values.
#Override
public boolean equals(Object o) {
if (o == this)
return true;
if (!(o instanceof EPoint)) {
return false;
}
EPoint ePoint = (EPoint) o;
return Math.abs(Math.abs(ePoint.lat) - Math.abs(lat)) < EPSILON && Math.abs(Math.abs(ePoint.lon) - Math.abs(lon)) < EPSILON;
}
#Override
public int hashCode() {
return Objects.hash(lat, lon);
}
I can't find a way to make the hasCode() consistent with my equals method.
Your equals itself breaks the contract even before you get to hashCode because it isn't transitive.
This also immediately leads to the only consistent hashCode implementation being to return a constant, because for any two points there is a (very long) chain of intermediate points so that
every two neighbors are equal, therefore
every two neighbors must have the same hashCode, therefore
beginning and end must have the same hashCode.
Now, this is a consistent implementation, but quite obviously a useless one.
I agree with Kayaman: The way your equals methos is implemented, you can have three EPoints (pointA,pointB,and pointC) with:
pointA.equals(pointB) //true
pointA.equals(pointC) //true
pointB.equals(pointC) //false
And this is not allowed. Creating a method with another name might be a solution.
If, however, you need your "almost Equal" objects to have the same hashcode, you could try a different approach:
Map every EPoint to an EPoint out of a grid. If, e.g. your EPoint's lat and lon where floats, you could map each EPoint to an EPoint with the rounded int-values.
If you need higher precision, you could extend on that and go into first, second,...decimal place).
If you do the equals() and the hashcode() method against the "mapped" Point, this should satisfy all requirements:
#Override
public boolean equals(Object o) {
if (o == this)
return true;
if (!(o instanceof EPoint)) {
return false;
}
EPoint ePoint = (EPoint) o;
return this.gridLon() == ePoint.gridLon() && ePoint.gridLat() == this.gridLat();
}
#Override
public int hashCode() {
return Objects.hash(this.gridLon(), this.gridLat());
}
Related
Does somebody know why this test fails? These are 2 identical objects with content inside, if I add the toString method after each Pair the test will pass. I tried to override the Equals method from the Object class but still nothing. Any help will be great, thanks
#Override
public int hashCode() {
return Objects.hash(key, value);
}
#Override
public boolean equals(Object obj) {
return super.equals(obj);
}
Pair Class Code
Error assertEquals
They are actual identical
Your problem is that your override of equals doesn't do anything - it just calls the implementation that you overrode. In other words, it has no effect.
You need to override equals in a way that does the comparison that you need. In other words, your equals needs to
check that the Object being compared to is also a Pair,
call equals for the two key fields,
call equals for the two value fields.
If any of these conditions fail, your equals should return false.
#Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Pair<?, ?> pair = (Pair<?, ?>) o;
return key.equals(pair.key) && value.equals(pair.value);
}
This code resolved problem :)
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.
I have a custom equals to check the equality of my object called Pair.
class Pair implements Comparable <Parr> {
double coef;
int power;
Pair(double a, int b) {
coef = a;
power = b;
}
My custom equals method is (located in class pair):
#Override
public boolean equals(Object o) {
if (!(o instanceof Pair))
return false;
Pair that = (Pair) o;
return that.coef == this.coef && that.power == this.power;
}
I've checked with print my object if the objects are the same, and they are indeed the same.
1.0 1 2.0 0
1.0 1 2.0 0
I call my custom equals from a different file, called Test.
class Test {
public static void main(String[] args) {
orig = pol1.differentiate().integrate();
System.out.print(orig);
if (orig.equals(pol1))
System.out.println(" (is equal.)");
else
System.out.println(" (is not equal.)");
And my class Polynomial, which is an arraylist with objects of Pair inside.
class Polynominal implements PolynominalInterface {
ArrayList<Pair> terms = new ArrayList<Pair>();
I looked on the internet, and I found that I cannot use == in my Equals method, but I'm using Intergers and Doubles, so equals() would not work.
Can anyone point me in the right direction?
If orig and pol1 are instances of Polynomial then this
if (orig.equals(pol1))
would only work if you implement Polynomial#equals() as well; which would iterate the two ArrayLists and make sure individual Pairs are equal (using Pair#equals() of course).
Ok, thanks to Ravi Thapliyal I found the solution.
After adding an custom equals method in my Polynominal class, the problem was fixed.
#Override
public boolean equals(Object o) {
if (!(o instanceof Polynomial))
return false;
Polynomial that = (Polynomial) o;
return that.terms.equals(terms);
}
Use the Double.compare(double, double) method instead of ==.
Floating point comparison is "fuzzy" in Java.
You would need to implement a Polynomail.equals() method something like the following:
public boolean equals(Object o) {
if (!(o instanceof Polynomial)) return false;
Polynomial other = (Polynomial) o;
if (this.terms==null && other.terms==null) return true;
// A suitable equals() method already exists for ArrayList, so we can use that
// this will in turn use Pair.equals() which looks OK to me
if (this.terms!=null && other.terms!=null) return this.terms.equals(other.terms);
return false;
}
Two issues come to mind: the first is that the default hashCode() method will seldom return the same value for any two distinct object instances, regardless of their contents. This is a good thing if the equals() method will never report two distinct object instances as equal, but is a bad thing if it will. Every object which overrides Object.equals() should also override Object.hashCode() so that if x.equals(y), then x.hashCode()==y.hashCode(); this is important because even non-hashed generic collections may use objects' hash codes to expedite comparisons. If you don't want to write a "real" hash function, simply pick some arbitrary integer and have your type's hashCode() method always return that. Any hashed collection into which your type is stored will perform slowly, but all collections into which it is stored should behave correctly.
The second issue you may be seeing is that floating-point comparisons are sometimes dodgy. Two numbers may be essentially equal but compare unequal. Worse, the IEEE decided for whatever reason that floating-point "not-a-number" values should compare unequal to everything--even themselves.
Factoring both of these issues together, I would suggest that you might want to rewrite your equals method to chain to the equals method of double. Further, if neither field of your object will be modified while it's stored in a collection, have your hashCode() method compute the hashCode of the int, multiply it by some large odd number, and then add or xor that with the hashCode of the double. If your object might be modified while stored in a collection, have hashCode() return a constant. If you don't override hashCode() you cannot expect the equals methods of any objects which contain yours to work correctly.
I'm wondering what collection I should use for this purpose:
Requirements
Must contain tuples <value1,value2>
There is not relation between those values (no key-value pairs)
Can only contain unique tuples
<value1,value2> is equal to <value2,value1>
What would be best to use here?
Use any Set (HashSet, for instance). Create an object to represent your tuple and implement hashcode and equals properly.
Implement your own tuple class with equals and hashCode as outlined below, then use Set:
public class Tuple<T> {
T v1;
T v2;
#override
public boolean equals(Object o) {
if (o == null) {
return false;
}
if (o instanceof Tuple) {
return (v1.equals(o.v1) && v2.equals(o.v2))
|| (v1.equals(o.v2) && v2.equals(o.v1));
}
return false;
}
#override
public int hashCode() {
// must produce a.hashCode() == b.hashCode() if a.equals(b)
// example below may or may not work for your concrete equals()
return v1.hashCode() ^ v2.hashCode();
}
}
A Set appears to meet your criteria. The Set would have to contain another collection or a custom object that contains the two values.