Related
Why the code is not compiling? so what is the reason why it is not compiling can someone state or elaborate it for me?
class Pet{
public Pet(int age){
System.out.print("Pet age: " + age);
}
}
public class Cat extends Pet{
public Cat(){
System.out.print("Cat");
}
public static void main(String[] args) {
new Pet(5);
}
}
Error Log:
Cat.java:7: error: constructor Pet in class Pet cannot be applied to given types;
public Cat(){
^
required: int
found: no arguments
reason: actual and formal argument lists differ in length
1 error
The Cat constructor needs to call the constructor of the superclass Pet with an age argument.
For example something like this:
public Cat(int age){
super(age)
System.out.print("Cat");
}
The error clearly says that you dont have default constructor in your parent class.
Every Java class has an empty constructor by default. but when you create a Parameterised constructor by yourself, It replaces the empty constructor.
To fix the error, either create an empty constructor in your Parent class, Or Call the parameterised constructor from your parent class.
This is the modified version of your code that should work.
class Pet{
public Pet(int age){
System.out.print("Pet age: " + age);
}
}
public class Cat extends Pet{
public Cat(int age){
super(age);
System.out.print("Cat");
}
public static void main(String[] args) {
new Cat(5);
}
}
Your Cat class inherits from the Pet class.
Pet class requires an integer in its default constructor.
Therefore the default constructor of Cat must call the Pet constructor with an integer parameter.
Alternatively, you can define a default constructor in Pet which does not take in an integer.
You should be able to call the parent constructor as such:
public class Cat extends Pet{
public Cat(){
super(3);
System.out.print("Cat");
}
}
When call is made to the Cat() constructor of Cat class, it will internally call the default constructor of parent class Pet. While when you declared the parameterized constructor for Pet class, it has no more default constructor exist there.
In order to fix it, you need to explicitly declare the default constructor for parent class Pet. Or explicitly call parameterized constructor of parent class.
In Cat constructor parent constructor(Pet) will be called. So you either you need to defined default no argument constructor in Pet class or call super(5) in Cat default constructor.
Default constructor missing in Pet.since you are extending Pet in Cat you need a super constructor to avoid this
class Pet{
public Pet(){};
public Pet(int age){
System.out.print("Pet age: " + age);
}
}
public class Cat extends Pet{
public Cat(){
System.out.print("Cat");
}
public static void main(String[] args) {
new Pet(5);
}
}
you have to call super class with an integer parameter. when main method call new object it calls Cat constructor it will search super class no arg constructor. but there is no no arg constructor so either need to added super class default constructor or call the super with parameter. sample code is below.
class Pet{
public Pet(int age){
System.out.print("Pet age: " + age)}}
public class Cat extends Pet{
public Cat(){
super(5);
System.out.print("Cat");}
public static void main(String[] args) {
new Pet(5);}}
Is it possible to call a constructor from another (within the same class, not from a subclass)? If yes how? And what could be the best way to call another constructor (if there are several ways to do it)?
Yes, it is possible:
public class Foo {
private int x;
public Foo() {
this(1);
}
public Foo(int x) {
this.x = x;
}
}
To chain to a particular superclass constructor instead of one in the same class, use super instead of this. Note that you can only chain to one constructor, and it has to be the first statement in your constructor body.
See also this related question, which is about C# but where the same principles apply.
Using this(args). The preferred pattern is to work from the smallest constructor to the largest.
public class Cons {
public Cons() {
// A no arguments constructor that sends default values to the largest
this(madeUpArg1Value,madeUpArg2Value,madeUpArg3Value);
}
public Cons(int arg1, int arg2) {
// An example of a partial constructor that uses the passed in arguments
// and sends a hidden default value to the largest
this(arg1,arg2, madeUpArg3Value);
}
// Largest constructor that does the work
public Cons(int arg1, int arg2, int arg3) {
this.arg1 = arg1;
this.arg2 = arg2;
this.arg3 = arg3;
}
}
You can also use a more recently advocated approach of valueOf or just "of":
public class Cons {
public static Cons newCons(int arg1,...) {
// This function is commonly called valueOf, like Integer.valueOf(..)
// More recently called "of", like EnumSet.of(..)
Cons c = new Cons(...);
c.setArg1(....);
return c;
}
}
To call a super class, use super(someValue). The call to super must be the first call in the constructor or you will get a compiler error.
[Note: I just want to add one aspect, which I did not see in the other answers: how to overcome limitations of the requirement that this() has to be on the first line).]
In Java another constructor of the same class can be called from a constructor via this(). Note however that this has to be on the first line.
public class MyClass {
public MyClass(double argument1, double argument2) {
this(argument1, argument2, 0.0);
}
public MyClass(double argument1, double argument2, double argument3) {
this.argument1 = argument1;
this.argument2 = argument2;
this.argument3 = argument3;
}
}
That this has to appear on the first line looks like a big limitation, but you can construct the arguments of other constructors via static methods. For example:
public class MyClass {
public MyClass(double argument1, double argument2) {
this(argument1, argument2, getDefaultArg3(argument1, argument2));
}
public MyClass(double argument1, double argument2, double argument3) {
this.argument1 = argument1;
this.argument2 = argument2;
this.argument3 = argument3;
}
private static double getDefaultArg3(double argument1, double argument2) {
double argument3 = 0;
// Calculate argument3 here if you like.
return argument3;
}
}
When I need to call another constructor from inside the code (not on the first line), I usually use a helper method like this:
class MyClass {
int field;
MyClass() {
init(0);
}
MyClass(int value) {
if (value<0) {
init(0);
}
else {
init(value);
}
}
void init(int x) {
field = x;
}
}
But most often I try to do it the other way around by calling the more complex constructors from the simpler ones on the first line, to the extent possible. For the above example
class MyClass {
int field;
MyClass(int value) {
if (value<0)
field = 0;
else
field = value;
}
MyClass() {
this(0);
}
}
Within a constructor, you can use the this keyword to invoke another constructor in the same class. Doing so is called an explicit constructor invocation.
Here's another Rectangle class, with a different implementation from the one in the Objects section.
public class Rectangle {
private int x, y;
private int width, height;
public Rectangle() {
this(1, 1);
}
public Rectangle(int width, int height) {
this( 0,0,width, height);
}
public Rectangle(int x, int y, int width, int height) {
this.x = x;
this.y = y;
this.width = width;
this.height = height;
}
}
This class contains a set of constructors. Each constructor initializes some or all of the rectangle's member variables.
As everybody already have said, you use this(…), which is called an explicit constructor invocation.
However, keep in mind that within such an explicit constructor invocation statement you may not refer to
any instance variables or
any instance methods or
any inner classes declared in this class or any superclass, or
this or
super.
As stated in JLS (§8.8.7.1).
Yes, any number of constructors can be present in a class and they can be called by another constructor using this() [Please do not confuse this() constructor call with this keyword]. this() or this(args) should be the first line in the constructor.
Example:
Class Test {
Test() {
this(10); // calls the constructor with integer args, Test(int a)
}
Test(int a) {
this(10.5); // call the constructor with double arg, Test(double a)
}
Test(double a) {
System.out.println("I am a double arg constructor");
}
}
This is known as constructor overloading.
Please note that for constructor, only overloading concept is applicable and not inheritance or overriding.
Using this keyword we can call one constructor in another constructor within same class.
Example :-
public class Example {
private String name;
public Example() {
this("Mahesh");
}
public Example(String name) {
this.name = name;
}
}
Yes it is possible to call one constructor from another. But there is a rule to it. If a call is made from one constructor to another, then
that new constructor call must be the first statement in the current constructor
public class Product {
private int productId;
private String productName;
private double productPrice;
private String category;
public Product(int id, String name) {
this(id,name,1.0);
}
public Product(int id, String name, double price) {
this(id,name,price,"DEFAULT");
}
public Product(int id,String name,double price, String category){
this.productId=id;
this.productName=name;
this.productPrice=price;
this.category=category;
}
}
So, something like below will not work.
public Product(int id, String name, double price) {
System.out.println("Calling constructor with price");
this(id,name,price,"DEFAULT");
}
Also, in the case of inheritance, when sub-class's object is created, the super class constructor is first called.
public class SuperClass {
public SuperClass() {
System.out.println("Inside super class constructor");
}
}
public class SubClass extends SuperClass {
public SubClass () {
//Even if we do not add, Java adds the call to super class's constructor like
// super();
System.out.println("Inside sub class constructor");
}
}
Thus, in this case also another constructor call is first declared before any other statements.
I will tell you an easy way
There are two types of constructors:
Default constructor
Parameterized constructor
I will explain in one Example
class ConstructorDemo
{
ConstructorDemo()//Default Constructor
{
System.out.println("D.constructor ");
}
ConstructorDemo(int k)//Parameterized constructor
{
this();//-------------(1)
System.out.println("P.Constructor ="+k);
}
public static void main(String[] args)
{
//this(); error because "must be first statement in constructor
new ConstructorDemo();//-------(2)
ConstructorDemo g=new ConstructorDemo(3);---(3)
}
}
In the above example I showed 3 types of calling
this() call to this must be first statement in constructor
This is Name less Object. this automatically calls the default constructor.
3.This calls the Parameterized constructor.
Note:
this must be the first statement in the constructor.
You can a constructor from another constructor of same class by using "this" keyword.
Example -
class This1
{
This1()
{
this("Hello");
System.out.println("Default constructor..");
}
This1(int a)
{
this();
System.out.println("int as arg constructor..");
}
This1(String s)
{
System.out.println("string as arg constructor..");
}
public static void main(String args[])
{
new This1(100);
}
}
Output -
string as arg constructor..
Default constructor..
int as arg constructor..
Calling constructor from another constructor
class MyConstructorDemo extends ConstructorDemo
{
MyConstructorDemo()
{
this("calling another constructor");
}
MyConstructorDemo(String arg)
{
System.out.print("This is passed String by another constructor :"+arg);
}
}
Also you can call parent constructor by using super() call
There are design patterns that cover the need for complex construction - if it can't be done succinctly, create a factory method or a factory class.
With the latest java and the addition of lambdas, it is easy to create a constructor which can accept any initialization code you desire.
class LambdaInitedClass {
public LamdaInitedClass(Consumer<LambdaInitedClass> init) {
init.accept(this);
}
}
Call it with...
new LambdaInitedClass(l -> { // init l any way you want });
Pretty simple
public class SomeClass{
private int number;
private String someString;
public SomeClass(){
number = 0;
someString = new String();
}
public SomeClass(int number){
this(); //set the class to 0
this.setNumber(number);
}
public SomeClass(int number, String someString){
this(number); //call public SomeClass( int number )
this.setString(someString);
}
public void setNumber(int number){
this.number = number;
}
public void setString(String someString){
this.someString = someString;
}
//.... add some accessors
}
now here is some small extra credit:
public SomeOtherClass extends SomeClass {
public SomeOtherClass(int number, String someString){
super(number, someString); //calls public SomeClass(int number, String someString)
}
//.... Some other code.
}
Hope this helps.
Yes it is possible to call one constructor from another with use of this()
class Example{
private int a = 1;
Example(){
this(5); //here another constructor called based on constructor argument
System.out.println("number a is "+a);
}
Example(int b){
System.out.println("number b is "+b);
}
You can call another constructor via the this(...) keyword (when you need to call a constructor from the same class) or the super(...) keyword
(when you need to call a constructor from a superclass).
However, such a call must be the first statement of your constructor. To overcome this limitation, use this answer.
The keyword this can be used to call a constructor from a constructor, when writing several constructor for a class, there are times when you'd like to call one constructor from another to avoid duplicate code.
Bellow is a link that I explain other topic about constructor and getters() and setters() and I used a class with two constructors. I hope the explanations and examples help you.
Setter methods or constructors
I know there are so many examples of this question but what I found I am putting here to share my Idea. there are two ways to chain constructor. In Same class you can use this keyword. in Inheritance, you need to use super keyword.
import java.util.*;
import java.lang.*;
class Test
{
public static void main(String args[])
{
Dog d = new Dog(); // Both Calling Same Constructor of Parent Class i.e. 0 args Constructor.
Dog cs = new Dog("Bite"); // Both Calling Same Constructor of Parent Class i.e. 0 args Constructor.
// You need to Explicitly tell the java compiler to use Argument constructor so you need to use "super" key word
System.out.println("------------------------------");
Cat c = new Cat();
Cat caty = new Cat("10");
System.out.println("------------------------------");
// Self s = new Self();
Self ss = new Self("self");
}
}
class Animal
{
String i;
public Animal()
{
i = "10";
System.out.println("Animal Constructor :" +i);
}
public Animal(String h)
{
i = "20";
System.out.println("Animal Constructor Habit :"+ i);
}
}
class Dog extends Animal
{
public Dog()
{
System.out.println("Dog Constructor");
}
public Dog(String h)
{
System.out.println("Dog Constructor with habit");
}
}
class Cat extends Animal
{
public Cat()
{
System.out.println("Cat Constructor");
}
public Cat(String i)
{
super(i); // Calling Super Class Paremetrize Constructor.
System.out.println("Cat Constructor with habit");
}
}
class Self
{
public Self()
{
System.out.println("Self Constructor");
}
public Self(String h)
{
this(); // Explicitly calling 0 args constructor.
System.out.println("Slef Constructor with value");
}
}
It is called Telescoping Constructor anti-pattern or constructor chaining. Yes, you can definitely do. I see many examples above and I want to add by saying that if you know that you need only two or three constructor, it might be ok. But if you need more, please try to use different design pattern like Builder pattern. As for example:
public Omar(){};
public Omar(a){};
public Omar(a,b){};
public Omar(a,b,c){};
public Omar(a,b,c,d){};
...
You may need more. Builder pattern would be a great solution in this case. Here is an article, it might be helpful
https://medium.com/#modestofiguereo/design-patterns-2-the-builder-pattern-and-the-telescoping-constructor-anti-pattern-60a33de7522e
Yes, you can call constructors from another constructor. For example:
public class Animal {
private int animalType;
public Animal() {
this(1); //here this(1) internally make call to Animal(1);
}
public Animal(int animalType) {
this.animalType = animalType;
}
}
you can also read in details from
Constructor Chaining in Java
Originally from an anser by Mirko Klemm, slightly modified to address the question:
Just for completeness: There is also the Instance initialization block that gets executed always and before any other constructor is called. It consists simply of a block of statements "{ ... }" somewhere in the body of your class definition. You can even have more than one. You can't call them, but they're like "shared constructor" code if you want to reuse some code across constructors, similar to calling methods.
So in your case
{
System.out.println("this is shared constructor code executed before the constructor");
field1 = 3;
}
There is also a "static" version of this to initialize static members: "static { ... }"
I prefer this way:
class User {
private long id;
private String username;
private int imageRes;
public User() {
init(defaultID,defaultUsername,defaultRes);
}
public User(String username) {
init(defaultID,username, defaultRes());
}
public User(String username, int imageRes) {
init(defaultID,username, imageRes);
}
public User(long id, String username, int imageRes) {
init(id,username, imageRes);
}
private void init(long id, String username, int imageRes) {
this.id=id;
this.username = username;
this.imageRes = imageRes;
}
}
There are two classes. One inherited from the other one. Test results surprise me
since I called the child class constructor and its method. However I do not call the super class constructor like this new Cougar().go();. But output shows the the super class constructor is called. Could you explain me why this happens?
public class Feline {
public String type = "f";
public Feline() {
System.out.println("feline");
}
}
public class Cougar extends Feline {
public Cougar() {
System.out.println("cougar");
}
public static void main(String[] args) {
new Cougar().go();
}
private void go() {
type = "c";
System.out.println(this.type + " " + super.type);
}
}
Output: feline
cougar
c c
If a constructor does not explicitly invoke a superclass constructor, the Java compiler automatically inserts a call to the no-argument constructor of the superclass.
The constructor of the superclass is always invoked. You can call the desired constructor of your superclass using "super()" in the first line of your child class constructor.
For example:
public class Feline {
public Feline() {
System.out.println("Feline");
}
public Feline(String type) {
System.out.println(type);
}
...
}
public class Cougar extends Feline{
public Cougar() {
super("Cougar");
}
...
}
It is Because of line
new Cougar().go();
Inside the "go" method you have redefined the variable "type".
I hope it is making sense.
When we invoke child class constructor,the java compiler automatically calls super class
constructor.If it is not present in super class,we will get compile time error.
If Object is the super class of our class then no issue since Object contains no-arg constructor.
Reason- The reason to have super constructor called is that if super class could have private fields which need to be initialized by its constructor.
Refer this link for more info.
When you call parameter-less constructor of the derived class the run-time automatically calls the parameter-less base class constructor before executing the derived class constructor. Because of this behavior you see feline cougar as first two string in your output.
Now for the 'c c' part of the output, you are not defining new String type variable in your Cougar class. You are using the same 'type' variable if the super class in go() method/function and setting it to "c". This is the reason you are getting 'c c' in the output.
Make Cougar as below. The syntax is C#.
public class Cougar : Feline
{
public String type = "c";
public Cougar()
{
Console.Write("cougar ");
}
static void Main(String[] args)
{
new Cougar().go();
}
private void go()
{
Console.Write(this.type + " " + base.type);
}
}
Good day!
I am reviewing the Java OO concept.
And wrote the following codes:
public class Main {
public static void main(String[] args) {
Animal dog = new Dog();
dog.eat();
dog.sleep();
}
}
abstract public class Animal {
private int age;
public Animal (){
age = 1;
}
public Animal (int age){ //How can I call this constructor?
this.age = age;
}
public void eat(){
System.out.println("Eat");
}
abstract public void sleep();
}
abstract public class Canine extends Animal{
abstract public void roam();
}
public interface Pet {
public String petName = null; //i want the pets to have a variable petName.
public void trick();
}
public class Dog extends Canine implements Pet{
public void roam(){
System.out.println("roam");
};
public void sleep(){
System.out.println("sleep");
};
public void eat(){
System.out.println("Eat Dog");
};
public void trick(){
System.out.println("trick");
}
}
I have several questions as follows:
How can I call the Animal Overloaded constructor?
How can I use the variable petName in the PET Interface?
Am I doing the concept of OO correctly? What rules am I violating?
Thank you in advance.
Subclasses will call the super constructor from within their own constructor using super(...) as the first line!
Interfaces cannot have variables or state - only methods!
You have a sound concept, but your code would not compile (because of item 2 above).
Some solutions:
public interface Pet {
String getName();
void trick();
}
Now the Dog class (or any class that implements Pet) will have to implement Pet.getName(). Give the Dog class a field of type String called 'name' and return it from Dog.getName().
public abstract class Canine extends Animal {
public Canine(int age) {
super(age); // pass the age parameter up to Animal
}
...
}
public class Dog extends Canine implements Pet {
private final String name;
public Dog(String name,int age) {
super(age); // call the age constructor
this.name=name;
}
public String getName() { return name; }
... rest of class ...
}
Each subclass (esp. the abstract ones) will need to provide matching constructors for all parent class constructors you want to call! (So I added the age parameter to the Canine constructor so that Dog could pass an age argument to it.
check out this ,it might help you,
http://www.roseindia.net/java/beginners/constructoroverloading.shtml
http://www.jchq.net/certkey/0602certkey.htm
Calling either constructor is calling an overloaded constructor: both constructors use the same name. To call the constructor that takes an int, call dog = new Dog(1);.
The class Dog implements Pet, so it will have a public field petName (is it super.petName?).
I don't see any fundamental errors.
1) you can call argumented contructor as" this(20);"
Example of explicit this constructor call
public class Point {
int mx;
int my;
//============ Constructor
public Point(int x, int y) {
mx = x;
my = y;
}
//============ Parameterless default constructor
public Point() {
this(0, 0); // Calls other constructor.
}
. . .
}super(...) - The superclass (parent) constructor
An object has the fields of its own class plus all fields of its parent class, grandparent class, all the way up to the root class Object. It's necessary to initialize all fields, therefore all constructors must be called! The Java compiler automatically inserts the necessary constructor calls in the process of constructor chaining, or you can do it explicitly.
The Java compiler inserts a call to the parent constructor (super) if you don't have a constructor call as the first statement of you constructor. The following is the equivalent of the constuctor above.
//============ Constructor (same as in above example)
public Point(int x, int y) {
super(); // Automatically done if you don't call constructor here.
m_x = x;
m_y = y;
}Why you might want to call super explicitly
Normally, you won't need to call the constructor for your parent class because it's automatically generated, but there are two cases where this is necessary.
You want to call a parent constructor which has parameters (the automatically generated super constructor call has no parameters).
There is no parameterless parent constructor because only constructors with parameters are defined in the parent class.
A call to the parent constructor is super(); in your case it would be super(12); for example. Note that a call to a constructor (either this() or super() must be the first statement. Note that you can only go up one level, so to call the constructor in Animal, you'd have to code that into Canine and not Dog.
public Dog(int age)
{
super(age); //this will invoke Canine's overloaded constructor
}
//You must now provide the overloaded constructor in Canine which invokes
//the overloaded constructor in Animal
public Canine(int age)
{
super(age); //this will invoke Animal's overloaded constructor
}
You may access the value of Pet.petName as, well, Pet.petName. This is because Interfaces can only have methods (which are implicitly public) and constants (which are implicitly public static final). You could use this.petName but it's pointless (and potentially confusing). You cannot set the value of petName anywhere else, ever since it's final, even if you don't declare it to be.
However in this case, declaring the name as part of the interface doesn't make sense since it's not constant. It should be a part of some abstract or concrete class so that it can be set per instance of Pet. You should instead define a getter (and maybe a setter) for Pet -
public String getName();
public String setName(String name);
these will force implementing classes to provide some sort of implementation to get and set the name for a Pet without enforcing how that name is stored inside of that class.
How can I call the Animal Overloaded constructor?
You need to chain it.
abstract public class Canine extends Animal{
public Canine(int age) {
super(age);
}
// ...
}
and
public class Dog extends Canine implements Pet{
public Dog(int age) {
super(age);
}
//...
}
How can I use the variable petName in the PET Interface?
Pet.petName. If a class implements such an interface, then the class can refer to those constants without a qualifying class name like this: petName
Am I doing the concept of OO correctly? What rules am I violating?
Interface constants
Is super() used to call the parent constructor?
Please explain super().
super() calls the parent constructor with no arguments.
It can be used also with arguments. I.e. super(argument1) and it will call the constructor that accepts 1 parameter of the type of argument1 (if exists).
Also it can be used to call methods from the parent. I.e. super.aMethod()
More info and tutorial here
Some facts:
super() is used to call the immediate parent.
super() can be used with instance members, i.e., instance variables and instance methods.
super() can be used within a constructor to call the constructor of the parent class.
OK, now let’s practically implement these points of super().
Check out the difference between program 1 and 2. Here, program 2 proofs our first statement of super() in Java.
Program 1
class Base
{
int a = 100;
}
class Sup1 extends Base
{
int a = 200;
void Show()
{
System.out.println(a);
System.out.println(a);
}
public static void main(String[] args)
{
new Sup1().Show();
}
}
Output:
200
200
Now check out program 2 and try to figure out the main difference.
Program 2
class Base
{
int a = 100;
}
class Sup2 extends Base
{
int a = 200;
void Show()
{
System.out.println(super.a);
System.out.println(a);
}
public static void main(String[] args)
{
new Sup2().Show();
}
}
Output:
100
200
In program 1, the output was only of the derived class. It couldn't print the variable of neither the base class nor the parent class. But in program 2, we used super() with variable a while printing its output, and instead of printing the value of variable a of the derived class, it printed the value of variable a of the base class. So it proves that super() is used to call the immediate parent.
OK, now check out the difference between program 3 and program 4.
Program 3
class Base
{
int a = 100;
void Show()
{
System.out.println(a);
}
}
class Sup3 extends Base
{
int a = 200;
void Show()
{
System.out.println(a);
}
public static void Main(String[] args)
{
new Sup3().Show();
}
}
Output:
200
Here the output is 200. When we called Show(), the Show() function of the derived class was called. But what should we do if we want to call the Show() function of the parent class? Check out program 4 for the solution.
Program 4
class Base
{
int a = 100;
void Show()
{
System.out.println(a);
}
}
class Sup4 extends Base
{
int a = 200;
void Show()
{
super.Show();
System.out.println(a);
}
public static void Main(String[] args)
{
new Sup4().Show();
}
}
Output:
100
200
Here we are getting two outputs, 100 and 200. When the Show() function of the derived class is invoked, it first calls the Show() function of the parent class, because inside the Show() function of the derived class, we called the Show() function of the parent class by putting the super keyword before the function name.
Source article: Java: Calling super()
Yes. super(...) will invoke the constructor of the super-class.
Illustration:
Stand alone example:
class Animal {
public Animal(String arg) {
System.out.println("Constructing an animal: " + arg);
}
}
class Dog extends Animal {
public Dog() {
super("From Dog constructor");
System.out.println("Constructing a dog.");
}
}
public class Test {
public static void main(String[] a) {
new Dog();
}
}
Prints:
Constructing an animal: From Dog constructor
Constructing a dog.
Is super() is used to call the parent constructor?
Yes.
Pls explain about Super().
super() is a special use of the super keyword where you call a parameterless parent constructor. In general, the super keyword can be used to call overridden methods, access hidden fields or invoke a superclass's constructor.
Here's the official tutorial
Calling the no-arguments super constructor is just a waste of screen space and programmer time. The compiler generates exactly the same code, whether you write it or not.
class Explicit() {
Explicit() {
super();
}
}
class Implicit {
Implicit() {
}
}
Yes, super() (lowercase) calls a constructor of the parent class. You can include arguments: super(foo, bar)
There is also a super keyword, that you can use in methods to invoke a method of the superclass
A quick google for "Java super" results in this
That is correct. Super is used to call the parent constructor. So suppose you have a code block like so
class A{
int n;
public A(int x){
n = x;
}
}
class B extends A{
int m;
public B(int x, int y){
super(x);
m = y;
}
}
Then you can assign a value to the member variable n.
I have seen all the answers. But everyone forgot to mention one very important point:
super() should be called or used in the first line of the constructor.
Just super(); alone will call the default constructor, if it exists of a class's superclass. But you must explicitly write the default constructor yourself. If you don't a Java will generate one for you with no implementations, save super(); , referring to the universal Superclass Object, and you can't call it in a subclass.
public class Alien{
public Alien(){ //Default constructor is written out by user
/** Implementation not shown…**/
}
}
public class WeirdAlien extends Alien{
public WeirdAlien(){
super(); //calls the default constructor in Alien.
}
}
For example, in selenium automation, you have a PageObject which can use its parent's constructor like this:
public class DeveloperSteps extends ScenarioSteps {
public DeveloperSteps(Pages pages) {
super(pages);
}........
I would like to share with codes whatever I understood.
The super keyword in java is a reference variable that is used to refer parent class objects. It is majorly used in the following contexts:-
1. Use of super with variables:
class Vehicle
{
int maxSpeed = 120;
}
/* sub class Car extending vehicle */
class Car extends Vehicle
{
int maxSpeed = 180;
void display()
{
/* print maxSpeed of base class (vehicle) */
System.out.println("Maximum Speed: " + super.maxSpeed);
}
}
/* Driver program to test */
class Test
{
public static void main(String[] args)
{
Car small = new Car();
small.display();
}
}
Output:-
Maximum Speed: 120
Use of super with methods:
/* Base class Person */
class Person
{
void message()
{
System.out.println("This is person class");
}
}
/* Subclass Student */
class Student extends Person
{
void message()
{
System.out.println("This is student class");
}
// Note that display() is only in Student class
void display()
{
// will invoke or call current class message() method
message();
// will invoke or call parent class message() method
super.message();
}
}
/* Driver program to test */
class Test
{
public static void main(String args[])
{
Student s = new Student();
// calling display() of Student
s.display();
}
}
Output:-
This is student class
This is person class
3. Use of super with constructors:
class Person
{
Person()
{
System.out.println("Person class Constructor");
}
}
/* subclass Student extending the Person class */
class Student extends Person
{
Student()
{
// invoke or call parent class constructor
super();
System.out.println("Student class Constructor");
}
}
/* Driver program to test*/
class Test
{
public static void main(String[] args)
{
Student s = new Student();
}
}
Output:-
Person class Constructor
Student class Constructor
What can we use SUPER for?
Accessing Superclass Members
If your method overrides some of its superclass's methods, you can invoke the overridden method through the use of the keyword super like super.methodName();
Invoking Superclass Constructors
If a constructor does not explicitly invoke a superclass constructor, the Java compiler automatically inserts a call to the no-argument constructor of the superclass. If the super class does not have a no-argument constructor, you will get a compile-time error.
Look at the code below:
class Creature {
public Creature() {
system.out.println("Creature non argument constructor.");
}
}
class Animal extends Creature {
public Animal (String name) {
System.out.println("Animal one argument constructor");
}
public Animal (Stirng name,int age) {
this(name);
system.out.println("Animal two arguments constructor");
}
}
class Wolf extends Animal {
public Wolf() {
super("tigerwang",33);
system.out.println("Wolf non argument constructor");
}
public static void main(string[] args) {
new Wolf();
}
}
When creating an object,the JVM always first execute the constructor in the class
of the top layer in the inheritance tree.And then all the way down the inheritance tree.The
reason why this is possible to happen is that the Java compiler automatically inserts a call
to the no-argument constructor of the superclass.If there's no non-argument constructor
in the superclass and the subclass doesn't explicitly say which of the constructor is to
be executed in the superclass,you'll get a compile-time error.
In the above code,if we want to create a Wolf object successfully,the constructor of the
class has to be executed.And during that process,the two-argu-constructor in the Animal
class is invoked.Simultaneously,it explicitly invokes the one-argu-constructor in the same
class and the one-argu-constructor implicitly invokes the non-argu-constructor in the Creature
class and the non-argu-constructor again implicitly invokes the empty constructor in the Object
class.
Constructors
In a constructor, you can use it without a dot to call another constructor. super calls a constructor in the superclass; this calls a constructor in this class :
public MyClass(int a) {
this(a, 5); // Here, I call another one of this class's constructors.
}
public MyClass(int a, int b) {
super(a, b); // Then, I call one of the superclass's constructors.
}
super is useful if the superclass needs to initialize itself. this is useful to allow you to write all the hard initialization code only once in one of the constructors and to call it from all the other, much easier-to-write constructors.
Methods
In any method, you can use it with a dot to call another method. super.method() calls a method in the superclass; this.method() calls a method in this class :
public String toString() {
int hp = this.hitpoints(); // Calls the hitpoints method in this class
// for this object.
String name = super.name(); // Calls the name method in the superclass
// for this object.
return "[" + name + ": " + hp + " HP]";
}
super is useful in a certain scenario: if your class has the same method as your superclass, Java will assume you want the one in your class; super allows you to ask for the superclass's method instead. this is useful only as a way to make your code more readable.
The super keyword can be used to call the superclass constructor and to refer to a member of the superclass
When you call super() with the right arguments, we actually call the constructor Box, which initializes variables width, height and depth, referred to it by using the values of the corresponding parameters. You only remains to initialize its value added weight. If necessary, you can do now class variables Box as private. Put down in the fields of the Box class private modifier and make sure that you can access them without any problems.
At the superclass can be several overloaded versions constructors, so you can call the method super() with different parameters. The program will perform the constructor that matches the specified arguments.
public class Box {
int width;
int height;
int depth;
Box(int w, int h, int d) {
width = w;
height = h;
depth = d;
}
public static void main(String[] args){
HeavyBox heavy = new HeavyBox(12, 32, 23, 13);
}
}
class HeavyBox extends Box {
int weight;
HeavyBox(int w, int h, int d, int m) {
//call the superclass constructor
super(w, h, d);
weight = m;
}
}
super is a keyword. It is used inside a sub-class method definition to call a method defined in the superclass. Private methods of the superclass cannot be called. Only public and protected methods can be called by the super keyword. It is also used by class constructors to invoke constructors of its parent class.
Check here for further explanation.
As stated, inside the default constructor there is an implicit super() called on the first line of the constructor.
This super() automatically calls a chain of constructors starting at the top of the class hierarchy and moves down the hierarchy .
If there were more than two classes in the class hierarchy of the program, the top class default constructor would get called first.
Here is an example of this:
class A {
A() {
System.out.println("Constructor A");
}
}
class B extends A{
public B() {
System.out.println("Constructor B");
}
}
class C extends B{
public C() {
System.out.println("Constructor C");
}
public static void main(String[] args) {
C c1 = new C();
}
}
The above would output:
Constructor A
Constructor B
Constructor C
The super keyword in Java is a reference variable that is used to refer to the immediate parent class object.
Usage of Java super Keyword
super can be used to refer to the immediate parent class instance variable.
super can be used to invoke the immediate parent class method.
super() can be used to invoke immediate parent class constructor.
There are a couple of other uses.
Referencing a default method of an inherited interface:
import java.util.Collection;
import java.util.stream.Stream;
public interface SkipFirstCollection<E> extends Collection<E> {
#Override
default Stream<E> stream() {
return Collection.super.stream().skip(1);
}
}
There is also a rarely used case where a qualified super is used to provide an outer instance to the superclass constructor when instantiating a static subclass:
public class OuterInstance {
public static class ClassA {
final String name;
public ClassA(String name) {
this.name = name;
}
public class ClassB {
public String getAName() {
return ClassA.this.name;
}
}
}
public static class ClassC extends ClassA.ClassB {
public ClassC(ClassA a) {
a.super();
}
}
public static void main(String[] args) {
final ClassA a = new ClassA("jeff");
final ClassC c = new ClassC(a);
System.out.println(c.getAName());
}
}
Then:
$ javac OuterInstance.java && java OuterInstance
jeff