Skip to main content

How to create Immutable Class or Object in Java? | Benefits of immutable class

Create Immutable Class in Java | Advantages of immutable class in java

How to create Immutable Class or Object in Java? | Benefits of immutable class

What is immutable class or object?

In simple term, once the object of the class created its fields cannot be changed or modified. Like once Student object is created, it can not be changed.

In java, all wrapper classes are immutable. like Boolean, String, Integer, Double, Long, Float.

Why we need Immutable class? or Advantages of Immutable class

  • When we does not want to change its fields after created.
  • It is Thread safe.
  • These objects are good for use as hash key (HashMap)
  • Reference of immutable objects can be cached.

How to create Immutable Class in Java?

We can create Immutable class using simple following steps :

  1. Make class Final
  2. Make class fields Private and Final
  3. Do not write Setter method for fields  
  4. Create all arguments constructor

Using above steps we can create Immutable class in Java. So lets go on code...

// Final class
final class Company {
    
    // Private fields
    final private int id;
    final private String name;

    // All arguments Constructor
    public Company(int id, String name) {
        this.id = id;
        this.name = name;
    }

    // Only getter methods
    public int getId() {
        return id;
    }
    
    public String getName() {
        return name;
    }
    
    @Override
    public String toString() {
        return "Company [id=" + id + ", name=" + name + "]";
    }

}

As above class follows all conditions we mentioned above, so we can call it is immutable class. But in this class if we add new field it can be mutable again lets see how we can solve that also.

Now lets create one Address class and add address field into Company class.

Address.java

public class Address {

    // Private fields
    final private int id;
    final private String location;
    
    public Address(int id, String location) {
        this.id = id;
        this.location = location;
    }
    
    public int getId() {
        return id;
    }
    
    public String getLocation() {
        return location;
    }
    
    @Override
    public String toString() {
        return "Address [id=" + id + ", location=" + location + "]";
    }
    
}

Company.java

// Final class
final class Company {
    
    // Private fields
    final private int id;
    final private String name;
    final private Address address;

    // All arguments Constructor
    public Company(int id, String name, Address address) {
        this.id = id;
        this.name = name;
        this.address = address;
    }

    // Only getter methods
    public int getId() {
        return id;
    }
    
    public String getName() {
        return name;
    }

    public Address getAddress() {
        return address;
    }

    @Override
    public String toString() {
        return "Company [id=" + id + ", name=" + name + ", address=" + address + "]";
    }
    
}

Demo.java

public class Demo {

    public static void main(String[] args) {
       
        Address addressObj = new Address(1, "India");
        Company companyObj = new Company(1, "Java", addressObj);

        System.out.println(companyObj);
       
        Address newAddress = companyObj.getAddress();
        newAddress.setLocation("USA");
       
        System.out.println(companyObj);
    }
}

Output :

Company [id=1, name=Java, address=Address [id=1, location=India]]
Company [id=1, name=Java, address=Address [id=1, location=USA]]

As you can see Address field changed in second output line and in immutable class we can not change any field. So how to fix that? lets see...

Step 1 : Return new Address in getAddress() method in Company class

Company.java

    public Address getAddress() {
        return new Address(address);
    }

Step 2 : Create new constructor in Address class and return address data

Address.java   

    public Address(Address address) {
        this(address.getId(), address.getLocation());
    } 

Output :

Company [id=1, name=Java, address=Address [id=1, location=India]]
Company [id=1, name=Java, address=Address [id=1, location=India]]

Now Address is not changing because every time it gives new object of Address.

Now Company class is immutable class.


Happy learning. Happy coding.

Other articles :



Comments

Popular posts from this blog

Flipping the Matrix HackerRank Solution in Java with Explanation

Java Solution for Flipping the Matrix | Find Highest Sum of Upper-Left Quadrant of Matrix Problem Description : Sean invented a game involving a 2n * 2n matrix where each cell of the matrix contains an integer. He can reverse any of its rows or columns any number of times. The goal of the game is to maximize the sum of the elements in the n *n submatrix located in the upper-left quadrant of the matrix. Given the initial configurations for q matrices, help Sean reverse the rows and columns of each matrix in the best possible way so that the sum of the elements in the matrix's upper-left quadrant is maximal.  Input : matrix = [[1, 2], [3, 4]] Output : 4 Input : matrix = [[112, 42, 83, 119], [56, 125, 56, 49], [15, 78, 101, 43], [62, 98, 114, 108]] Output : 119 + 114 + 56 + 125 = 414 Full Problem Description : Flipping the Matrix Problem Description   Here we can find solution using following pattern, So simply we have to find Max of same number of box like (1,1,1,1). And ...

Sales by Match HackerRank Solution | Java Solution

HackerRank Sales by Match problem solution in Java   Problem Description : Alex works at a clothing store. There is a large pile of socks that must be paired by color for sale. Given an array of integers representing the color of each sock, determine how many pairs of socks with matching colors there are. For example, there are n=7 socks with colors socks = [1,2,1,2,1,3,2]. There is one pair of color 1 and one of color 2 . There are three odd socks left, one of each color. The number of pairs is 2 .   Example 1 : Input : n = 6 arr = [1, 2, 3, 4, 5, 6] Output : 0 Explanation : We have 6 socks with all different colors, So print 0. Example 2 : Input : n = 10 arr = [1, 2, 3, 4, 1, 4, 2, 7, 9, 9] Output : 4 Explanation : We have 10 socks. There is pair of color 1, 2, 4 and 9, So print 4. This problem easily solved by HashMap . Store all pair of socks one by one in Map and check if any pair is present in Map or not. If pair is present then increment ans variable by 1 ...