Related
I am trying to override equals method in Java. I have a class People which basically has 2 data fields name and age. Now I want to override equals method so that I can check between 2 People objects.
My code is as follows
public boolean equals(People other){
boolean result;
if((other == null) || (getClass() != other.getClass())){
result = false;
} // end if
else{
People otherPeople = (People)other;
result = name.equals(other.name) && age.equals(other.age);
} // end else
return result;
} // end equals
But when I write age.equals(other.age) it gives me error as equals method can only compare String and age is Integer.
Solution
I used == operator as suggested and my problem is solved.
//Written by K#stackoverflow
public class Main {
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
ArrayList<Person> people = new ArrayList<Person>();
people.add(new Person("Subash Adhikari", 28));
people.add(new Person("K", 28));
people.add(new Person("StackOverflow", 4));
people.add(new Person("Subash Adhikari", 28));
for (int i = 0; i < people.size() - 1; i++) {
for (int y = i + 1; y <= people.size() - 1; y++) {
boolean check = people.get(i).equals(people.get(y));
System.out.println("-- " + people.get(i).getName() + " - VS - " + people.get(y).getName());
System.out.println(check);
}
}
}
}
//written by K#stackoverflow
public class Person {
private String name;
private int age;
public Person(String name, int age){
this.name = name;
this.age = age;
}
#Override
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (obj.getClass() != this.getClass()) {
return false;
}
final Person other = (Person) obj;
if ((this.name == null) ? (other.name != null) : !this.name.equals(other.name)) {
return false;
}
if (this.age != other.age) {
return false;
}
return true;
}
#Override
public int hashCode() {
int hash = 3;
hash = 53 * hash + (this.name != null ? this.name.hashCode() : 0);
hash = 53 * hash + this.age;
return hash;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}
Output:
run:
-- Subash Adhikari - VS - K false
-- Subash Adhikari - VS - StackOverflow false
-- Subash Adhikari - VS - Subash Adhikari true
-- K - VS - StackOverflow false
-- K - VS - Subash Adhikari false
-- StackOverflow - VS - Subash Adhikari false
-- BUILD SUCCESSFUL (total time: 0 seconds)
Introducing a new method signature that changes the parameter types is called overloading:
public boolean equals(People other){
Here People is different than Object.
When a method signature remains the identical to that of its superclass, it is called overriding and the #Override annotation helps distinguish the two at compile-time:
#Override
public boolean equals(Object other){
Without seeing the actual declaration of age, it is difficult to say why the error appears.
I'm not sure of the details as you haven't posted the whole code, but:
remember to override hashCode() as well
the equals method should have Object, not People as its argument type. At the moment you are overloading, not overriding, the equals method, which probably isn't what you want, especially given that you check its type later.
you can use instanceof to check it is a People object e.g. if (!(other instanceof People)) { result = false;}
equals is used for all objects, but not primitives. I think you mean age is an int (primitive), in which case just use ==. Note that an Integer (with a capital 'I') is an Object which should be compared with equals.
See What issues should be considered when overriding equals and hashCode in Java? for more details.
Item 10: Obey the general contract when overriding equals
According to Effective Java, Overriding the equals method seems simple, but there are many ways to get it wrong, and consequences can be dire. The easiest way to avoid problems is not to override the equals method, in which case each instance of the class is equal only to itself. This is the right thing to do if any of the following conditions apply:
Each instance of the class is inherently unique. This is true for classes such as Thread that represent active entities rather than values. The equals implementation provided by Object has exactly the right behavior for these classes.
There is no need for the class to provide a “logical equality” test. For example, java.util.regex.Pattern could have overridden equals to check whether two Pattern instances represented exactly the same regular expression, but the designers didn’t think that clients would need or want this functionality. Under these circumstances, the equals implementation inherited from Object is ideal.
A superclass has already overridden equals, and the superclass behavior is appropriate for this class. For example, most Set implementations inherit their equals implementation from AbstractSet, List implementations from AbstractList, and Map implementations from AbstractMap.
The class is private or package-private, and you are certain that its equals method will never be invoked. If you are extremely risk-averse, you can override the equals method to ensure that it isn’t invoked accidentally:
The equals method implements an equivalence relation. It has these properties:
Reflexive: For any non-null reference value x, x.equals(x) must return true.
Symmetric: For any non-null reference values x and y, x.equals(y) must return true if and only if y.equals(x) returns true.
Transitive: For any non-null reference values x, y, z, if x.equals(y) returns true and y.equals(z) returns true, then x.equals(z) must return true.
Consistent: For any non-null reference values x and y, multiple invocations of x.equals(y) must consistently return true or consistently return false, provided no information used in equals comparisons is modified.
For any non-null reference value x, x.equals(null) must return false.
Here’s a recipe for a high-quality equals method:
Use the == operator to check if the argument is a reference to this object. If so, return true. This is just a performance optimization but one that is worth doing if the comparison is potentially expensive.
Use the instanceof operator to check if the argument has the correct type. If not, return false. Typically, the correct type is the class in which the method occurs. Occasionally, it is some interface implemented by this class. Use an interface if the class implements an interface that refines the equals contract to permit comparisons across classes that implement the interface. Collection interfaces such as Set, List, Map, and Map.Entry have this property.
Cast the argument to the correct type. Because this cast was preceded by an instanceof test, it is guaranteed to succeed.
For each “significant” field in the class, check if that field of the argument matches the corresponding field of this object. If all these tests succeed, return true; otherwise, return false. If the type in Step 2 is an interface, you must access the argument’s fields via interface methods; if the type is a class, you may be able to access the fields directly, depending on their accessibility.
For primitive fields whose type is not float or double, use the == operator for comparisons; for object reference fields, call the equals method recursively; for float fields, use the static Float.compare(float, float) method; and for double fields, use Double.compare(double, double). The special treatment of float and double fields is made necessary by the existence of Float.NaN, -0.0f and the analogous double values; While you could compare float and double fields with the static methods Float.equals and Double.equals, this would entail autoboxing on every comparison, which would have poor performance. For array fields, apply these guidelines to each element. If every element in an array field is significant, use one of the Arrays.equals methods.
Some object reference fields may legitimately contain null. To avoid the possibility of a NullPointerException, check such fields for equality using the static method Objects.equals(Object, Object).
// Class with a typical equals method
public final class PhoneNumber {
private final short areaCode, prefix, lineNum;
public PhoneNumber(int areaCode, int prefix, int lineNum) {
this.areaCode = rangeCheck(areaCode, 999, "area code");
this.prefix = rangeCheck(prefix, 999, "prefix");
this.lineNum = rangeCheck(lineNum, 9999, "line num");
}
private static short rangeCheck(int val, int max, String arg) {
if (val < 0 || val > max)
throw new IllegalArgumentException(arg + ": " + val);
return (short) val;
}
#Override public boolean equals(Object o) {
if (o == this)
return true;
if (!(o instanceof PhoneNumber))
return false;
PhoneNumber pn = (PhoneNumber)o;
return pn.lineNum == lineNum && pn.prefix == prefix
&& pn.areaCode == areaCode;
}
... // Remainder omitted
}
#Override
public boolean equals(Object that){
if(this == that) return true;//if both of them points the same address in memory
if(!(that instanceof People)) return false; // if "that" is not a People or a childclass
People thatPeople = (People)that; // than we can cast it to People safely
return this.name.equals(thatPeople.name) && this.age == thatPeople.age;// if they have the same name and same age, then the 2 objects are equal unless they're pointing to different memory adresses
}
When comparing objects in Java, you make a semantic check, comparing the type and identifying state of the objects to:
itself (same instance)
itself (clone, or reconstructed copy)
other objects of different types
other objects of the same type
null
Rules:
Symmetry: a.equals(b) == b.equals(a)
equals() always yields true or false, but never a NullpointerException, ClassCastException or any other throwable
Comparison:
Type check: both instances need to be of the same type, meaning you have to compare the actual classes for equality. This is often not correctly implemented, when developers use instanceof for type comparison (which only works as long as there are no subclasses, and violates the symmetry rule when A extends B -> a instanceof b != b instanceof a).
Semantic check of identifying state: Make sure you understand by which state the instances are identified. Persons may be identified by their social security number, but not by hair color (can be dyed), name (can be changed) or age (changes all the time). Only with value objects should you compare the full state (all non-transient fields), otherwise check only what identifies the instance.
For your Person class:
public boolean equals(Object obj) {
// same instance
if (obj == this) {
return true;
}
// null
if (obj == null) {
return false;
}
// type
if (!getClass().equals(obj.getClass())) {
return false;
}
// cast and compare state
Person other = (Person) obj;
return Objects.equals(name, other.name) && Objects.equals(age, other.age);
}
Reusable, generic utility class:
public final class Equals {
private Equals() {
// private constructor, no instances allowed
}
/**
* Convenience equals implementation, does the object equality, null and type checking, and comparison of the identifying state
*
* #param instance object instance (where the equals() is implemented)
* #param other other instance to compare to
* #param stateAccessors stateAccessors for state to compare, optional
* #param <T> instance type
* #return true when equals, false otherwise
*/
public static <T> boolean as(T instance, Object other, Function<? super T, Object>... stateAccessors) {
if (instance == null) {
return other == null;
}
if (instance == other) {
return true;
}
if (other == null) {
return false;
}
if (!instance.getClass().equals(other.getClass())) {
return false;
}
if (stateAccessors == null) {
return true;
}
return Stream.of(stateAccessors).allMatch(s -> Objects.equals(s.apply(instance), s.apply((T) other)));
}
}
For your Person class, using this utility class:
public boolean equals(Object obj) {
return Equals.as(this, obj, t -> t.name, t -> t.age);
}
Since I'm guessing age is of type int:
public boolean equals(Object other){
boolean result;
if((other == null) || (getClass() != other.getClass())){
result = false;
} // end if
else{
People otherPeople = (People)other;
result = name.equals(otherPeople.name) && age == otherPeople.age;
} // end else
return result;
} // end equals
if age is int you should use == if it is Integer object then you can use equals().
You also need to implement hashcode method if you override equals. Details of the contract is available in the javadoc of Object and also at various pages in web.
tl;dr
record Person ( String name , int age ) {}
if(
new Person( "Carol" , 27 ) // Compiler auto-generates implicitly the constructor.
.equals( // Compiler auto-generates implicitly the `equals` method.
new Person( "Carol" , 42 )
)
) // Returns `false`, as the name matches but the age differs.
{ … }
Details
While your specific problem is solved (using == for equality test between int primitive values), there is an alternative that eliminates the need to write that code.
record
Java 16 brings the record feature.
A record is a brief way to write a class whose main purpose is to transparently and immutably carry data. The compiler implicitly creates the constructor, getters, equals & hashCode, and toString.
equals method provided automatically
The default implicit equals method compares each and every member field that you declared for the record. The members can be objects or primitives, both types are automatically compared in the default equals method.
For example, if you have a Person record carrying two fields, name & age, both of those fields are automatically compared to determine equality between a pair of Person objects.
public record Person ( String name , int age ) {}
Try it.
Person alice = new Person( "Alice" , 23 ) ;
Person alice2 = new Person( "Alice" , 23 ) ;
Person bob = new Person( "Bob" , 19 ) ;
boolean samePerson1 = alice.equals( alice2 ) ; // true.
boolean samePerson2 = alice.equals( bob ) ; // false.
You can override the equals method on a record, if you want a behavior other than the default. But if you do override equals, be sure to override hashCode for consistent logic, as you would for a conventional Java class. And, think twice: Whenever adding methods to a record, reconsider if a record structure is really appropriate to that problem domain.
Tip: A record can be defined within another class, and even locally within a method.
Here is the solution that I recently used:
public class Test {
public String a;
public long b;
public Date c;
public String d;
#Override
public boolean equals(Object obj) {
if (this == obj) {
return true;
}
if (!(obj instanceof Test)) {
return false;
}
Test testOther = (Test) obj;
return (a != null ? a.equals(testOther.a) : testOther.a == null)
&& (b == testOther.b)
&& (c != null ? c.equals(testOther.c) : testOther.c == null)
&& (d != null ? d.equals(testOther.d) : testOther.d == null);
}
}
For lazy programmers: lombok library is very easy and time saving. please have a look at this link
instead of writing lines of codes and rules, you just need to apply this library in your IDE and then just #Data and it is Done.
import lombok.Data;
#Data // this is the magic word :D
public class pojo {
int price;
String currency;
String productName;
}
in fact in the above code, #Data is a shortcut for
import lombok.Data;
import lombok.EqualsAndHashCode;
import lombok.Getter;
import lombok.Setter;
import lombok.ToString;
#Getter
#Setter
#EqualsAndHashCode
#ToString
//or instead of all above #Data
public class pojo {
int price;
String currency;
String productName;
}
This question already has answers here:
Understanding the workings of equals and hashCode in a HashMap
(9 answers)
Closed 2 years ago.
I have a Test class and its objects with same value (101,"abc") is created twice. I am inserting these two object in Hashmap as a Key. I want to understand the internal functioning as to why I am getting size as two of a map when both of my keys are same, it should probably overwrite ?
import java.util.HashMap;
import java.util.Map;
public class Test{
int id;
String name;
public static void main(String[] args) {
Test e1 = new Test(101,"abc");
Test e2 = new Test(101,"abc");
//Test e3 = new Test(101,"abc");
Map<Test, String> map = new HashMap<>();
map.put(e1, "XYZ");
map.put(e2, "CDF");
String value = map.get(e2);
System.out.println( "VALUE : "+value);
}
public Test(int id, String name) {
this.id = id;
this.name=name;
}}
Test e1 = new Test(101,"abc");
Test e2 = new Test(101,"abc");
Creates 2 different objects of type Test. This means 2 different memory space has been allocated for e1 & e2.
Now lets understand how does map( say Hash Map) identifies the uniqueness of key( in much simpler words, how does the map knows that the key you are trying to insert is already present or not). Answer is, map calls the hashcode() & equals() method to compare the keys present in map with the one you are trying to insert.
As we already know all classes have a default parent class Object . And the Test class is not having an implementation of hashcode() & equals(); so when map tried calling them, the object class method were called.
As Object's class equals returns true only when both the objects in comparison refers to the same object reference & so does the hashcode() and here e1 & e2 are clearly different objects. So you got two entries.
Solution is to have an override the equals() & hashcode() as suggested by #Kapil above.
There will be two objects in the hashmap unless you override hashcode() and equals() methods in your Test class. If you override these two methods on the id property of Test class then the two keys with the same id would be treated as duplicate and you will see the expected behaviour.
Example below states that you want to make two object equals if their ids are equal -
#Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + id;
return result;
}
#Override
public boolean equals(Object obj) {
if (this == obj)
return true;
if (obj == null)
return false;
if (getClass() != obj.getClass())
return false;
Test other = (Test) obj;
if (id != other.id)
return false;
return true;
}
I am trying to override equals method in Java. I have a class People which basically has 2 data fields name and age. Now I want to override equals method so that I can check between 2 People objects.
My code is as follows
public boolean equals(People other){
boolean result;
if((other == null) || (getClass() != other.getClass())){
result = false;
} // end if
else{
People otherPeople = (People)other;
result = name.equals(other.name) && age.equals(other.age);
} // end else
return result;
} // end equals
But when I write age.equals(other.age) it gives me error as equals method can only compare String and age is Integer.
Solution
I used == operator as suggested and my problem is solved.
//Written by K#stackoverflow
public class Main {
/**
* #param args the command line arguments
*/
public static void main(String[] args) {
// TODO code application logic here
ArrayList<Person> people = new ArrayList<Person>();
people.add(new Person("Subash Adhikari", 28));
people.add(new Person("K", 28));
people.add(new Person("StackOverflow", 4));
people.add(new Person("Subash Adhikari", 28));
for (int i = 0; i < people.size() - 1; i++) {
for (int y = i + 1; y <= people.size() - 1; y++) {
boolean check = people.get(i).equals(people.get(y));
System.out.println("-- " + people.get(i).getName() + " - VS - " + people.get(y).getName());
System.out.println(check);
}
}
}
}
//written by K#stackoverflow
public class Person {
private String name;
private int age;
public Person(String name, int age){
this.name = name;
this.age = age;
}
#Override
public boolean equals(Object obj) {
if (obj == null) {
return false;
}
if (obj.getClass() != this.getClass()) {
return false;
}
final Person other = (Person) obj;
if ((this.name == null) ? (other.name != null) : !this.name.equals(other.name)) {
return false;
}
if (this.age != other.age) {
return false;
}
return true;
}
#Override
public int hashCode() {
int hash = 3;
hash = 53 * hash + (this.name != null ? this.name.hashCode() : 0);
hash = 53 * hash + this.age;
return hash;
}
public int getAge() {
return age;
}
public void setAge(int age) {
this.age = age;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
}
Output:
run:
-- Subash Adhikari - VS - K false
-- Subash Adhikari - VS - StackOverflow false
-- Subash Adhikari - VS - Subash Adhikari true
-- K - VS - StackOverflow false
-- K - VS - Subash Adhikari false
-- StackOverflow - VS - Subash Adhikari false
-- BUILD SUCCESSFUL (total time: 0 seconds)
Introducing a new method signature that changes the parameter types is called overloading:
public boolean equals(People other){
Here People is different than Object.
When a method signature remains the identical to that of its superclass, it is called overriding and the #Override annotation helps distinguish the two at compile-time:
#Override
public boolean equals(Object other){
Without seeing the actual declaration of age, it is difficult to say why the error appears.
I'm not sure of the details as you haven't posted the whole code, but:
remember to override hashCode() as well
the equals method should have Object, not People as its argument type. At the moment you are overloading, not overriding, the equals method, which probably isn't what you want, especially given that you check its type later.
you can use instanceof to check it is a People object e.g. if (!(other instanceof People)) { result = false;}
equals is used for all objects, but not primitives. I think you mean age is an int (primitive), in which case just use ==. Note that an Integer (with a capital 'I') is an Object which should be compared with equals.
See What issues should be considered when overriding equals and hashCode in Java? for more details.
Item 10: Obey the general contract when overriding equals
According to Effective Java, Overriding the equals method seems simple, but there are many ways to get it wrong, and consequences can be dire. The easiest way to avoid problems is not to override the equals method, in which case each instance of the class is equal only to itself. This is the right thing to do if any of the following conditions apply:
Each instance of the class is inherently unique. This is true for classes such as Thread that represent active entities rather than values. The equals implementation provided by Object has exactly the right behavior for these classes.
There is no need for the class to provide a “logical equality” test. For example, java.util.regex.Pattern could have overridden equals to check whether two Pattern instances represented exactly the same regular expression, but the designers didn’t think that clients would need or want this functionality. Under these circumstances, the equals implementation inherited from Object is ideal.
A superclass has already overridden equals, and the superclass behavior is appropriate for this class. For example, most Set implementations inherit their equals implementation from AbstractSet, List implementations from AbstractList, and Map implementations from AbstractMap.
The class is private or package-private, and you are certain that its equals method will never be invoked. If you are extremely risk-averse, you can override the equals method to ensure that it isn’t invoked accidentally:
The equals method implements an equivalence relation. It has these properties:
Reflexive: For any non-null reference value x, x.equals(x) must return true.
Symmetric: For any non-null reference values x and y, x.equals(y) must return true if and only if y.equals(x) returns true.
Transitive: For any non-null reference values x, y, z, if x.equals(y) returns true and y.equals(z) returns true, then x.equals(z) must return true.
Consistent: For any non-null reference values x and y, multiple invocations of x.equals(y) must consistently return true or consistently return false, provided no information used in equals comparisons is modified.
For any non-null reference value x, x.equals(null) must return false.
Here’s a recipe for a high-quality equals method:
Use the == operator to check if the argument is a reference to this object. If so, return true. This is just a performance optimization but one that is worth doing if the comparison is potentially expensive.
Use the instanceof operator to check if the argument has the correct type. If not, return false. Typically, the correct type is the class in which the method occurs. Occasionally, it is some interface implemented by this class. Use an interface if the class implements an interface that refines the equals contract to permit comparisons across classes that implement the interface. Collection interfaces such as Set, List, Map, and Map.Entry have this property.
Cast the argument to the correct type. Because this cast was preceded by an instanceof test, it is guaranteed to succeed.
For each “significant” field in the class, check if that field of the argument matches the corresponding field of this object. If all these tests succeed, return true; otherwise, return false. If the type in Step 2 is an interface, you must access the argument’s fields via interface methods; if the type is a class, you may be able to access the fields directly, depending on their accessibility.
For primitive fields whose type is not float or double, use the == operator for comparisons; for object reference fields, call the equals method recursively; for float fields, use the static Float.compare(float, float) method; and for double fields, use Double.compare(double, double). The special treatment of float and double fields is made necessary by the existence of Float.NaN, -0.0f and the analogous double values; While you could compare float and double fields with the static methods Float.equals and Double.equals, this would entail autoboxing on every comparison, which would have poor performance. For array fields, apply these guidelines to each element. If every element in an array field is significant, use one of the Arrays.equals methods.
Some object reference fields may legitimately contain null. To avoid the possibility of a NullPointerException, check such fields for equality using the static method Objects.equals(Object, Object).
// Class with a typical equals method
public final class PhoneNumber {
private final short areaCode, prefix, lineNum;
public PhoneNumber(int areaCode, int prefix, int lineNum) {
this.areaCode = rangeCheck(areaCode, 999, "area code");
this.prefix = rangeCheck(prefix, 999, "prefix");
this.lineNum = rangeCheck(lineNum, 9999, "line num");
}
private static short rangeCheck(int val, int max, String arg) {
if (val < 0 || val > max)
throw new IllegalArgumentException(arg + ": " + val);
return (short) val;
}
#Override public boolean equals(Object o) {
if (o == this)
return true;
if (!(o instanceof PhoneNumber))
return false;
PhoneNumber pn = (PhoneNumber)o;
return pn.lineNum == lineNum && pn.prefix == prefix
&& pn.areaCode == areaCode;
}
... // Remainder omitted
}
#Override
public boolean equals(Object that){
if(this == that) return true;//if both of them points the same address in memory
if(!(that instanceof People)) return false; // if "that" is not a People or a childclass
People thatPeople = (People)that; // than we can cast it to People safely
return this.name.equals(thatPeople.name) && this.age == thatPeople.age;// if they have the same name and same age, then the 2 objects are equal unless they're pointing to different memory adresses
}
When comparing objects in Java, you make a semantic check, comparing the type and identifying state of the objects to:
itself (same instance)
itself (clone, or reconstructed copy)
other objects of different types
other objects of the same type
null
Rules:
Symmetry: a.equals(b) == b.equals(a)
equals() always yields true or false, but never a NullpointerException, ClassCastException or any other throwable
Comparison:
Type check: both instances need to be of the same type, meaning you have to compare the actual classes for equality. This is often not correctly implemented, when developers use instanceof for type comparison (which only works as long as there are no subclasses, and violates the symmetry rule when A extends B -> a instanceof b != b instanceof a).
Semantic check of identifying state: Make sure you understand by which state the instances are identified. Persons may be identified by their social security number, but not by hair color (can be dyed), name (can be changed) or age (changes all the time). Only with value objects should you compare the full state (all non-transient fields), otherwise check only what identifies the instance.
For your Person class:
public boolean equals(Object obj) {
// same instance
if (obj == this) {
return true;
}
// null
if (obj == null) {
return false;
}
// type
if (!getClass().equals(obj.getClass())) {
return false;
}
// cast and compare state
Person other = (Person) obj;
return Objects.equals(name, other.name) && Objects.equals(age, other.age);
}
Reusable, generic utility class:
public final class Equals {
private Equals() {
// private constructor, no instances allowed
}
/**
* Convenience equals implementation, does the object equality, null and type checking, and comparison of the identifying state
*
* #param instance object instance (where the equals() is implemented)
* #param other other instance to compare to
* #param stateAccessors stateAccessors for state to compare, optional
* #param <T> instance type
* #return true when equals, false otherwise
*/
public static <T> boolean as(T instance, Object other, Function<? super T, Object>... stateAccessors) {
if (instance == null) {
return other == null;
}
if (instance == other) {
return true;
}
if (other == null) {
return false;
}
if (!instance.getClass().equals(other.getClass())) {
return false;
}
if (stateAccessors == null) {
return true;
}
return Stream.of(stateAccessors).allMatch(s -> Objects.equals(s.apply(instance), s.apply((T) other)));
}
}
For your Person class, using this utility class:
public boolean equals(Object obj) {
return Equals.as(this, obj, t -> t.name, t -> t.age);
}
Since I'm guessing age is of type int:
public boolean equals(Object other){
boolean result;
if((other == null) || (getClass() != other.getClass())){
result = false;
} // end if
else{
People otherPeople = (People)other;
result = name.equals(otherPeople.name) && age == otherPeople.age;
} // end else
return result;
} // end equals
if age is int you should use == if it is Integer object then you can use equals().
You also need to implement hashcode method if you override equals. Details of the contract is available in the javadoc of Object and also at various pages in web.
tl;dr
record Person ( String name , int age ) {}
if(
new Person( "Carol" , 27 ) // Compiler auto-generates implicitly the constructor.
.equals( // Compiler auto-generates implicitly the `equals` method.
new Person( "Carol" , 42 )
)
) // Returns `false`, as the name matches but the age differs.
{ … }
Details
While your specific problem is solved (using == for equality test between int primitive values), there is an alternative that eliminates the need to write that code.
record
Java 16 brings the record feature.
A record is a brief way to write a class whose main purpose is to transparently and immutably carry data. The compiler implicitly creates the constructor, getters, equals & hashCode, and toString.
equals method provided automatically
The default implicit equals method compares each and every member field that you declared for the record. The members can be objects or primitives, both types are automatically compared in the default equals method.
For example, if you have a Person record carrying two fields, name & age, both of those fields are automatically compared to determine equality between a pair of Person objects.
public record Person ( String name , int age ) {}
Try it.
Person alice = new Person( "Alice" , 23 ) ;
Person alice2 = new Person( "Alice" , 23 ) ;
Person bob = new Person( "Bob" , 19 ) ;
boolean samePerson1 = alice.equals( alice2 ) ; // true.
boolean samePerson2 = alice.equals( bob ) ; // false.
You can override the equals method on a record, if you want a behavior other than the default. But if you do override equals, be sure to override hashCode for consistent logic, as you would for a conventional Java class. And, think twice: Whenever adding methods to a record, reconsider if a record structure is really appropriate to that problem domain.
Tip: A record can be defined within another class, and even locally within a method.
Here is the solution that I recently used:
public class Test {
public String a;
public long b;
public Date c;
public String d;
#Override
public boolean equals(Object obj) {
if (this == obj) {
return true;
}
if (!(obj instanceof Test)) {
return false;
}
Test testOther = (Test) obj;
return (a != null ? a.equals(testOther.a) : testOther.a == null)
&& (b == testOther.b)
&& (c != null ? c.equals(testOther.c) : testOther.c == null)
&& (d != null ? d.equals(testOther.d) : testOther.d == null);
}
}
For lazy programmers: lombok library is very easy and time saving. please have a look at this link
instead of writing lines of codes and rules, you just need to apply this library in your IDE and then just #Data and it is Done.
import lombok.Data;
#Data // this is the magic word :D
public class pojo {
int price;
String currency;
String productName;
}
in fact in the above code, #Data is a shortcut for
import lombok.Data;
import lombok.EqualsAndHashCode;
import lombok.Getter;
import lombok.Setter;
import lombok.ToString;
#Getter
#Setter
#EqualsAndHashCode
#ToString
//or instead of all above #Data
public class pojo {
int price;
String currency;
String productName;
}
I have a obj type MyObj as follow:
class MyObj{
String id;
String username;
String fullName;
String age;
//getters & setters
}
Suppose we have 2 lists containing different number of elements like so:
List<MyObj> listA
List<MyObj> listB
I have a generic method that detects elements from listA that are missing in listB:
public static <T> List<T> getListDifference(List<T> list1, List<T> list2) {
Collection<T> first = new HashSet<T>(list1);
Collection<T> second = new HashSet<T>(list2);
first.removeAll(second);
return new ArrayList<T>(first);
}
If objects from listA and listB has the same fields for all items, everything works just fine.
The problem is that some items has only id and username but others can have fullName or age too, and as result this method doesn't work anymore. I'd like to keep the same logic, considering only id field because it's present in all objects.
One obvious method is to copy only object's id field in other List<String> and work with obtained lists to detect elements, then just search for obj from both lists by id. This method has a big complexity, because of multiple iterations. Is there a short way to achieve this?
As # Kevin Esche mentioned, you have to implement equals() and hashCode() in your MyObj POJO, and according to your situation, they should be:
#Override
public int hashCode() {
return Objects.hash(id);
}
#Override
public boolean equals(Object obj) {
if (obj == this) return true;
if (!(obj instanceof MyObj)) return false;
MyObj myObj= (MyObj) obj;
return Objects.equals(id, myObj.id);
}
That will work even if ages are different and the ids are the same for example.
override MyObjs equals method:
#Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
MyObj myObj = (MyObj) o;
return (id != null ? !id.equals(myObj.id) : myObj.id != null);
}
IntelliJ/Android Studio have helpers for automatically generating equals and hashCode where you can choose the necessary properties.
You have to implement equals and hashCode to work with sets. Sets needs a way of recognising if one object is equal to another, and default behaviour is not gonna work here.
By the way, I suggest using retainAll method on a list, check this out.
This question already has answers here:
Java HashMap return value not confirming with my understanding of equals and hashcode
(4 answers)
Closed 8 years ago.
What I know is: While inserting elements in the HashMap, Java checks value of hashCode and inserts that element in inside the HashMap and while retrieving the object from HashMap, Java checks the value of HashCode and retrieved the object that has the value generated from that HashCode. Is this correct?
I created a modal to override default implementation of HashCode. Every time that modal is called, it gives back a same value. So, if we add that modal again and again, why entries in HashMap are increasing?
Here is my code:
Modal:
public class MyModal {
int empId;
String empName;
#Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
MyModal myModal = (MyModal) o;
if (empId != myModal.empId) return false;
if (empName != null ? !empName.equals(myModal.empName) : myModal.empName != null) return false;
return true;
}
#Override
public int hashCode() {
return 1;
}
public MyModal(int empId, String empName) {
this.empId = empId;
this.empName = empName;
}
}
public class TestHashCode {
public static void main(String[] args) {
HashMap<MyModal, Integer> hashMap = new HashMap<>();
MyModal modal1 = new MyModal(1, "a");
MyModal modal2 = new MyModal(2, "b");
hashMap.put(modal1, 1);
hashMap.put(modal2, 2);
System.out.println("Size is" + hashMap.size());
System.out.println(modal1.hashCode() + " "+modal2.hashCode());
}
}
Output:
Size is2
1 1
Hash collisions aren't the only limiting factor in the map construction, however returning a constant 1 is a "worst" case Map and it behaves like a LinkedList (every element is in one bucket). From the Object.hashCode() Javadoc
It is not required that if two objects are unequal according to the equals(java.lang.Object) method, then calling the hashCode method on each of the two objects must produce distinct integer results. However, the programmer should be aware that producing distinct integer results for unequal objects may improve the performance of hash tables.
Both HashCode and Equals need to be implemented. Hash code is used to narrow the search in equals need to be implemented to define equality. Two unequal objects can have same hashcode but does not mean they are equal.
http://docs.oracle.com/javase/7/docs/api/java/lang/Object.html#hashCode().