Home > Java > javaTutorial > body text

Constructor overloading using static blocks in Java

王林
Release: 2023-09-06 14:41:06
forward
1347 people have browsed it

Constructor overloading using static blocks in Java

The act of instantiating an object calls its corresponding constructor, which is the basis for many features in object-oriented programming. It's worth noting that in any program that uses objects, there will invariably be a default constructor - the compiler automatically creates it for seamless use.

In this discussion, we will delve into constructor overloading of static blocks in Java. Constructor overloading is the concept of defining multiple constructors with different parameters in a class.

grammar

Public class class_name {
   Class_name() {
   }
   Class_name(par1, par2..) {
   }
}
Copy after login

Using constructors with static blocks provides more power and flexibility during object initialization.

algorithm

To use a static block to overload the constructor, please follow the steps below -

  • Step 1 - Create a class with multiple constructors with different parameters.

  • Step 2 - Create a static block using the "static" keyword

    This block is executed once when the class is loaded into memory.

  • Step 3- When loading a class, first the static block is executed and then the constructor is executed when the object is created.

  • Step 4 - The constructor will be called based on the arguments provided.

Method 1: Define a separate static block

This approach involves defining static blocks and overloaded constructors separately.

The Chinese translation of

Example

is:

Example

Class Class_name{
   Static {
   }
   Public class_name(){
   }
   Public class_name(int value) {
   }
   Public class_name(string name) {
   }
   //Other methods
}
Copy after login

In this approach, a class can have overloaded constructors with different parameter lists, which also include initialization code. There is also a separate static block for static initialization of the class. This block will be executed once.

Example

In this example, we will show method 1

class Emp { int id, exp;String name;static String company;
   static { company = "XYZ Corp"; }
   public Emp(){
      System.out.println("-"
                           + "\t"
                           + "-"
                           + "\t"
                           + "-"
                           + "\t"
                           + "\t"
                           + "-"); }
   public Emp(int id, String name){ System.out.println(id + "\t" + name + "\t"
                           + company + "\t" + exp); }
   public Emp(int id, String name, int exp) {
      System.out.println(id + "\t" + name + "\t"  + company + "\t" + exp); }
}
public class Way2Class {
   public static void main(String[] args) {
      System.out.println("Id"
                           + "\t"
                           + "Name"
                           + "\t"
                           + "Company"
                           + "\t"
                           + "Exp");
      Emp obj1 = new Emp(001, "Apoorva");
      Emp obj2 = new Emp(004, "Anu", 10);
      Emp obj3 = new Emp();
   }
}
Copy after login

Output

Id	Name	Company	Exp
1	Apoorva	XYZ Corp	0
4	Anu	XYZ Corp	10
-	-	-		-
Copy after login

illustrate

In a company, employees with any number of years of experience will work for the same company. So if no value is passed in company variable then it automatically sets the same company name as company. For this we use static blocks.

Method 2: Call static method from constructor

To perform shared initialization code, you can declare static methods in the class and call them from the constructor.

The Chinese translation of

Example

is:

Example

public class Way2Class {
   private int value;
   private String name;

   private static void initialize() {
      System.out.println("Common initialization code");
   }

   public Way2Class() {
      initialize();
      value = 0;
      name = "Default";
      System.out.println("No-arg constructor called");
   }

   public Way2Class(int value) {
      initialize();
      this.value = value;
      name = "Value";
      System.out.println("Int constructor called");
   }

   public Way2Class(String name) {
      initialize();
      value = 0;
      this.name = name;
      System.out.println("String constructor called");
   }

   public static void main(String[] args) {
      Way2Class obj1 = new Way2Class();
      Way2Class obj2 = new Way2Class(10);
      Way2Class obj3 = new Way2Class("Hello");
   }
}
Copy after login

Output

Common initialization code
No-arg constructor called
Common initialization code
Int constructor called
Common initialization code
String constructor called
Copy after login

illustrate

The Way2Class class in this picture contains three constructors, each of which calls the static initialize () method to execute shared initialization code. Each constructor calls the static function initialize() specified within the class. Based on the given parameters, the appropriate constructor is called during object creation, and the static method initialize() is used to execute the public initialization code.

Comparison between Method 1 and Method 2

standard

method 1

Method 2

type

Separate static block

Call static method from constructor

method

Reuse common static methods with different constructors.

Independent static methods and common constructors.

Method logic

Constructor overloading and static blocks

Constructor overloading and static blocks

in conclusion

While approach 2 (static method called from constructor) provides greater flexibility in code organization and inheritance, approach 1 (multiple constructors with common code) is more independent and simpler. The choice between the two methods depends on the specific requirements and design considerations of the current project.

The above is the detailed content of Constructor overloading using static blocks in Java. For more information, please follow other related articles on the PHP Chinese website!

Related labels:
source:tutorialspoint.com
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template