Develop Reference

Compact equals and hashcode

I have a bean with 4 attributes:
user
institutionId
groupId
postingDate
I use Eclipse to generate equals and hashcode but the resulting code is not pretty. Is there a compact way to do the same? Assuming I want equals & hashcode to use all the attributes or a subset of them.
#Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((groupId == null) ? 0 : groupId.hashCode());
result = prime * result + ((institutionId == null) ? 0 : institutionId.hashCode());
result = prime * result + ((postingDate == null) ? 0 : postingDate.hashCode());
result = prime * result + ((user == null) ? 0 : user.hashCode());
return result;
}
#Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
ManGroupKey other = (ManGroupKey) obj;
if (groupId == null) {
if (other.groupId != null)
return false;
} else if (!groupId.equals(other.groupId))
return false;
if (institutionId == null) {
if (other.institutionId != null)
return false;
} else if (!institutionId.equals(other.institutionId))
return false;
if (postingDate == null) {
if (other.postingDate != null)
return false;
} else if (!postingDate.equals(other.postingDate))
return false;
if (user == null) {
if (other.user != null)
return false;
} else if (!user.equals(other.user))
return false;
return true;
}

In Java 7
public int hashCode() {
return Objects.hash(groupId, institutionId, postingDate, user);
}
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
// cast to correct class
Target o = (Target)obj;
return Objects.equals(groupId, o.groupId) &&
Objects.equals(institutionId, o.institutionId) &&
Objects.equals(postingDate, o.postingDate) &&
Objects.equals(user, o.user);
}

You could compact the code down, but the odds are far higher that you would introduce bugs than that you would do anything useful. All the parts of the equals and hash code method are there for a reason.
If it's bothering you most IDEs have a folding editor, just click the little yellow box (usually) and all the contents of the method get hidden away.

Instead of using the eclipse generated code, you can use Apache-common-langs(http://commons.apache.org/proper/commons-lang/) class HashCodeBuilder and EqualsBuilder to do this:
public int hashCode() {
return HashCodeBuilder.reflectionHashCode(this);
}
public boolean equals(Object obj) {
return EqualsBuilder.reflectionEquals(this);
}

hashCode:
Either:
#Override
public int hashCode() {
return Objects.hash(user, institutionId, groupId, postingDate);
}
Or:
#Override
public int hashCode() {
int result = 17;
result = 31 * result + Objects.hashCode(user);
result = 31 * result + Objects.hashCode(institutionId);
result = 31 * result + Objects.hashCode(groupId);
result = 31 * result + Objects.hashCode(postingDate);
return result;
}
Equals:
public boolean equals(Object obj){
if (obj == this){
return true;
}
if (! (obj instanceof ManGroupKey)){
return false;
}
ManGroupKey other = (ManGroupKey) obj;
return Objects.equals(user, other.user)
&& Objects.equals(institutionId, other.institutionId)
&& Objects.equals(groupId, other.groupId)
&& Objects.equals(postingDate, other.postingDate);
}

You can at least remove one level of nesting by removing the other.x != null check.
Comparing a value in this way: x.equals(y) will always return false when y is null.
Aside from that: the .equals() method is a good example where a bit of reflection can be handy, possible extracted out into a generic utility method. All you have to do is run through the different fields and see if they're equal in the two objects, that can be done in a few lines.
Obviously that is only feasible when you actually want to compare each field (or you'll have to add some additions to it which let you choose the fields).

I think the library, that can suite you is apache common. It provides EqualsBuilder and HashCodeBuilder classes, that do exactly what you are looking for.
Consider this question for details: Apache Commons equals/hashCode builder
Here are some code snippets:
public class Bean{
private String name;
private int length;
private List<Bean> children;
#Override
public int hashCode(){
return new HashCodeBuilder()
.append(name)
.append(length)
.append(children)
.toHashCode();
}
#Override
public boolean equals(final Object obj){
if(obj instanceof Bean){
final Bean other = (Bean) obj;
return new EqualsBuilder()
.append(name, other.name)
.append(length, other.length)
.append(children, other.children)
.isEquals();
} else{
return false;
}
}
}

Comparing Fractions with Bool

So Im having a problem with my equals class
public boolean equals(Object other) {
if (other.equals(this.numerator) && other.equals(this.denominator))
return true;
else
return false;
}
it gives me the result of 9/2 eq 9/2= false.
(Rest of my code for referees )
https://gist.github.com/anonymous/6604f427cc9d17391478
What am i doing wrong?
I edited the code but still and dealing with an error of boolean and int
public boolean equals(Object other) {
if (other.equals(this.numerator) == getNumerator() && other.equals(this.denominator)== getDenominator())
return true;
else
return false;
}

When implementing equals, before checking if the objects are equal you should consider the next scenarios:
The two are actually references to the same object
The other object is null
The other object is an instance of a different type
and when checking equality between the two objects you should consider the nullness of every field involved in the comparison too. In most IDEs equals can be generated automatically, in eclipse:
Alt + Shift + s --> Generate hashCode() and equals()
the next is 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;
A other = (A) obj;
if (denominator == null) {
if (other.denominator != null)
return false;
} else if (!denominator.equals(other.denominator))
return false;
if (numerator == null) {
if (other.numerator != null)
return false;
} else if (!numerator.equals(other.numerator))
return false;
return true;
}

return
(other.numerator.equals(this.numerator)
&&
other.denominator.equals(this.denominator))
Of couse this doesn't consider that 1/2 == 2/4.

How do I change what it means for objects to be duplicate in a set?

I want to store a set of Edges:
class Edge {
int u;
int v;
char symbol;
}
The problem is that it's possible for two Edge objects to have the same u, v and symbol, but they can both be stored in a HashSet because they're not the same object even though I want them to be considered the same object. How can I store only one object that has a unique (u, v, symbol) in a Set?

You need to override the following two methods equals and hashcode.
public boolean equals(Object obj) {
if (obj == null) return false;
if (!(obj instanceof Edge)) return false;
// return true if they are the same, otherwise false
}
public int hashCode() {
// return an int that represents similarity
// Example: name.hashCode(), if they are the same with the same name
}

Depends on what kind of set you want to use; The below applies for HashSet for instance, but not for any subclass of SortedSet
By overriding equals() and hashCode():
class Edge {
int u;
int v;
char symbol;
#Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + symbol;
result = prime * result + u;
result = prime * result + v;
return result;
}
#Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Edge other = (Edge) obj;
return symbol == other.symbol && u == other.u && v == other.v;
}
}

You have to override equals(). Like this:
public boolean equals(Object obj) {
//do the comparison here; remember to cast obj to Edge
}

contains function linkedlist

public class state implements Comparator<state>{
Point a;
Point b;
private int path_cost=0;
...
}
class Point {
int x;
int y;
...
}
for above i have:
PriorityQueue<state> openNode= new PriorityQueue<state>();
LinkedList<state> closed =new LinkedList<state>();
state currNode;
I need to check if the Point a of ANY openNode or closed equals currNode's Point a.
I could use contains if i had to match the entire object but here i just care about one variabale (Point a) of state class. I want the method to check all the nodes in PriorityQueue and LinkedList.
addition:
I am thinking about using Iterator on my priorityQueue and LinkedList. But i am not sure how to read the value of Point a using Iterator.

EDIT: Looked like I'd misunderstood slightly. It's simpler than I thought.
// I've assumed more conventional names
Point currPoint = currNode.getPointA();
for (State openNode : openNodes) {
if (openNode.getPointA().equals(currPoint)) {
return true;
}
}
for (State closedNode : closedNodes) {
if (closedNode.getPointA().equals(currPoint)) {
return true;
}
}
// No matching points
return false;
You could potentially use Guava's Iterables.concat() method to make this slightly simpler:
for (State node : Iterables.concat(closedNodes, openNodes)) {
if (node.getPointA().equals(currPoint)) {
return true;
}
}
return false;
If you need to know which node has an equal point A, just change it to:
for (State node : Iterables.concat(closedNodes, openNodes)) {
if (node.getPointA().equals(currPoint)) {
return node;
}
}
return null;
That will only find one such node, of course - there may be multiple matches.

You will have to either provide equals method on Point a for state class or just use simple iteration and iterate over both List for comparison. contains method does the same.
If you use any other method it will be time consuming.
Very odd method is use Comparator to check equality
class PointAComparator implements Comparator<State>
{
Point p = null;
public PointAComparator(Point a) {
p = a;
}
#Override
public int compare(State o1, State o2) {
return (p.x == o1.a.x && p.y == o1.a.y) ? 1
: (p.x == o2.a.x && p.y == o2.a.y) ? 1 : -1;
}
}
Above compare method returns 1 for equal else -1 so when you do sorting then each list will have elements at the start which are equal. and then you can check for first element.

i used method overriding on function equals for both the object and achieved my result.
class Point {
int x;
int y;
...
#Override
public boolean equals(Object other){
if (other == null) return false;
if (other == this) return true;
if (!(other instanceof Point))return false;
Point otherPoint = (Point)other;
return (this.x==otherPoint.getX() && this.y==otherPoint.getY() )? true : false;
}
}
public class state implements Comparator<state>{
Point a;
Point b;
private int path_cost=0;
...
#Override
public boolean equals(Object other){
if (other == null) return false;
if (other == this) return true;
if (!(other instanceof state))return false;
state otherState = (state)other;
return ((this.a).equals(otherState.a))? true : false;
}
}

Override the equals method

I'm currently revising for an exam. On a past paper there was the question,
Override the equals method in the following class. The method shall check for
content equality of the whole state.
class Employee
{
String firstName;
int age;
}
[2 marks]
I did some fiddling for the right answer and have come up so far with this. Is there a simpler way to answer the question and is this right? Many thanks for help.
public class Employee
{
int age;
public boolean equals(Object obj)
{
if(this == obj)
{
return true; //Reference equality.
}
if(!(obj instanceof Employee))
{
return false; // not the same type.
}
Employee other = (Employee) obj;
return firstName == other.firstName;
return age == other.age;
}
}

Use
return (((this.firstName == null || other.firstName == null)
&& this.firstName == other.firstName)
||
this.firstName.equals(other.firstName))
&& age == other.age;
This handles null cases too.

One thing to note, and I wouldn't imagine you would get docked for this in an exam...
It's usually poor practice to do an instanceof when the class isn't final. The reason for that is that equals() must be symmetric. Accepting subclasses (who might also implement equals with their own new aspects) could cause it to not be symmetric.
Example (I think the example is the same used in Effective Java 2ed):
class Point {
protected int x, y;
//equals method uses instanceof Point and checks x and y values are the same
}
class ColorPoint extends Point {
protected Color color;
//equals method checks that it's a ColorPoint, that super.equals is true,
//then checks the Color
}
new Point(1, 2).equals(new ColorPoint(1, 2, Color.red)); //true
new ColorPoint(1, 2, Color.red).equals(new Point(1, 2)); //false
It's a very subtle point that even most of the answerers here didn't take into account. But it's the reason that most generators of equals (such as the one in your favourite IDE) tend to do exact class comparison:
if ( this.getClass() != other.getClass() ) {
return false;
}
When the equals method uses instanceof it's usually a good move to document that subclasses must follow the exact same specification.

A couple of points:
You need to check if obj is null.
To compare String contents in Java, use equals(), i.e. firstName.equals(other.firstName). Check to see if firstName is null first.
Here's an improved implementation:
public boolean equals(Object obj)
{
if(obj == null)
{
return false;
}
if(this == obj)
{
return true; //Reference equality.
}
if(this.getClass() != obj.getClass())
{
return false;
}
Employee other = (Employee) obj;
if(firstName == null)
{
if(other.firstName != null)
{
return false;
}
}
else if(!firstName.equals(other.firstName))
{
return false;
}
return age == other.age;
}
EDIT: Updated type comparison to make equals() symmetric in accordance with #Mark Peters' answer.

The String firstName should be compared with .equals(), NOT ==. The == compare is ok for the primitive int age field.
What if both firstNames are identical, yet age is unequal? Shouldn't this fail?
Something like return (firstName.equals(obj.firstName)) && (age == obj.age);
Of course, that doesn't consider the case when this.firstName is null, which would result in a NullPointerException being thrown.
Are the Employees considered equal if both have null firstNames? What if one is null and the other not? Assuming both must be null, or both must be String.equals(), you could use:
return ((null == firstName && null == obj.firstName)
|| (null != firstName && firstName.equals(obj.firstName)))
&& (age == obj.age);
instead of your 2 return statements. The rest looks ok.

public boolean equals(Object o){
if(this==o){ //same instance, no need to check more
return true;
}
if(o instanceof Employee){ //when null this will yield false
Employee other = (Employee) o;
return (this.name == other.name || (this.name != null && this.name.equals(other.name)) && this.age == other.age;
}
return false;
}

To put together in one answer all the parts already mentioned:
public boolean equals(Object obj) {
if(this == obj) {
return true; //Reference equality.
}
if(obj == null || !(obj instanceof Employee))
{
return false; // not the same type.
}
Employee other = (Employee) obj;
return (firstName.equals(other.firstName) && age == other.age);
}

Your last line, the age comparison, is unreachable; you shouldn't use == to compare Strings; and you need to account for null values.
Since the general movement seems to be towards laying it all out for you, here's Eclipse's implementation:
public class Employee {
private final String firstName;
private final int age;
public Employee(final String firstName, final int age) {
super();
this.firstName = firstName;
this.age = age;
}
#Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + age;
result = prime * result + ((firstName == null) ? 0 : firstName.hashCode());
return result;
}
#Override
public boolean equals(final Object obj) {
if (this == obj) {
return true;
}
if (obj == null) {
return false;
}
if (getClass() != obj.getClass()) {
return false;
}
final Employee other = (Employee) obj;
if (age != other.age) {
return false;
}
if (firstName == null) {
if (other.firstName != null) {
return false;
}
} else if (!firstName.equals(other.firstName)) {
return false;
}
return true;
}
}
And a battery of tests:
import org.junit.Test;
public class EmployeeTest {
#Test
public void testEmployeeEquals() {
final Employee nullNameEmp = new Employee(null, -1);
final Employee empA1 = new Employee("a", 1);
final Employee empA1Clone = new Employee("a", 1);
final Employee empA2 = new Employee("a", 2);
final Employee empB1 = new Employee("b", 1);
final Employee empB2 = new Employee("b", 2);
final Employee subEmp = new Employee("a", 1) {
};
assert !nullNameEmp.equals(empA1);
assert !nullNameEmp.equals(empA1Clone);
assert !nullNameEmp.equals(empA2);
assert !nullNameEmp.equals(empB1);
assert !nullNameEmp.equals(empB2);
assert !nullNameEmp.equals(subEmp);
assert !nullNameEmp.equals(null);
assert !empA1.equals(nullNameEmp);
assert empA1.equals(empA1Clone);
assert !empA1.equals(empA2);
assert !empA1.equals(empB1);
assert !empA1.equals(empB2);
assert !empA1.equals(subEmp);
assert !empA1.equals(null);
assert !empA2.equals(nullNameEmp);
assert !empA2.equals(empA1);
assert !nullNameEmp.equals(empA1Clone);
assert !empA2.equals(empB1);
assert !empA2.equals(empB2);
assert !empA2.equals(subEmp);
assert !empA2.equals(null);
assert !empB1.equals(nullNameEmp);
assert !empB1.equals(empA1);
assert !empB1.equals(empA1Clone);
assert !empB1.equals(empA2);
assert !empB1.equals(empB2);
assert !empB1.equals(subEmp);
assert !empB1.equals(null);
assert !empB2.equals(nullNameEmp);
assert !empB2.equals(empA1);
assert !empB2.equals(empA1Clone);
assert !empB2.equals(empA2);
assert !empB2.equals(empB1);
assert !empB2.equals(subEmp);
assert !empB2.equals(null);
assert !subEmp.equals(nullNameEmp);
assert !subEmp.equals(empA1);
assert !subEmp.equals(empA1Clone);
assert !subEmp.equals(empA2);
assert !subEmp.equals(empB1);
assert !subEmp.equals(empB2);
assert !subEmp.equals(null);
assert nullNameEmp.equals(nullNameEmp);
assert empA1.equals(empA1);
assert empA1Clone.equals(empA1Clone);
assert empA2.equals(empA2);
assert empB1.equals(empB1);
assert empB2.equals(empB2);
assert subEmp.equals(subEmp);
}
}