Overloading in java inheritance

Method Overloading in Java

When one class has the same method names but they differ by parameter declaration then methods are called overloaded and this mechanism is called method overloading. When an overloaded method is called, JVM determines which method to call by checking the type and number of the parameters. Let us understand this by the below example.

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class overloadDemo < void print() < System.out.println("0 Parameter"); >// Overload print for one integer parameter. void print(int i) < System.out.println("i:" + i); >void print(int i, int j) < System.out.println("i and j : " + i + " " + j); >// Overload print for a double parameter double print(double e) < System.out.println("e: " + e); return e*e; >> public class demo < public static void main(String args[]) < overloadDemo o = new overloadDemo(); double r; // call all versions of print() o.print(); o.print(5); o.print(15, 10); r = o.print(0.80); System.out.println("r : " + r); >>

0 Parameter i:5 i and j : 15 10 e: 0.8 r: 0.6400000000000001

Here we can see method print is overloaded with 4 different versions. The first version is not returning anything and is not taking any parameters as well. The second version is accepting an integer parameter without returning any value. The third version is accepting 2 integer parameters without returning anything. The fourth version is accepting a double type parameter and is returning a double value to the caller. Whenever the overloaded method is called, Java will look for the best match and call that method that has the same parameter type and return type.

Method Overloading in Java -Table of Contents

Constructor Overloading in Java with Example

Along with methods, Java also allows overloading of constructors. When any object is created, we can initialize it with different values. But when we don’t want to initialize it with any values or when we want to initialize only with one value, at that time we can use constructors overloading. Let us understand it in detail by the below example:

class boxDemo < double width; double height; double depth; boxDemo(double w, double h, double d) < width = w; height = h; depth = d; >boxDemo() < width = -1; height = -1; depth = -1; >boxDemo(double len) < width = height = depth = len; >double volume() < return width * height * depth; >> public class overloadDemo < public static void main(String args[]) < boxDemo ob1 = new boxDemo(5, 15, 25); boxDemo ob2 = new boxDemo(); boxDemo ob3 = new boxDemo(5); double v; v = ob1.volume(); System.out.println("volume of ob1 is " + v); v = ob2.volume(); System.out.println("volume of ob2 is " + v); v = ob3.volume(); System.out.println("volume of ob3 is " + v); >>

volume of ob1 is 1875.0 volume of ob2 is -1.0 volume of ob3 is 125.0

The respective constructor will be called based upon the parameters passed to it.

[ Related Article:- Java Tutorial ]

Recursion in Java

When a method calls itself then it is called a recursive method and this mechanism is called recursion. The best example of recursion is a factorial calculation of any number.

class factDemo < // this is a recursive method int fact(int n) < int r; if(n==1) return 1; r = fact(n-1) * n; return r; >> public class recursionDemo < public static void main(String args[]) < factDemo f = new factDemo(); System.out.println("Factorial of 2 is " + f.fact(2)); System.out.println("Factorial of 3 is " + f.fact(3)); System.out.println("Factorial of 4 is " + f.fact(4)); >>

Factorial of 2 is 2 Factorial of 3 is 6 Factorial of 4 is 24

Its main advantage is to develop the code of few algorithms in a simpler and clearer way.

Inheritance in java

Java provides the mechanism called inheritance where one class can acquire the properties and behavior of another class. We can group this concept into 2 below groups:

• superclass(parent): a class that is inherited
• subclass(child): class inheriting another class

A subclass inherits all the variables and methods of the superclass and adds its own unique properties. In Java, we use the keyword extends to inherit the class.

Let us understand through the below simple example.

class X < int i; void printi() < System.out.println("i: " + i ); >> // Create a subclass by extending class A. class Y extends X < int j; void printj() < System.out.println("j: " + j); >void sum() < System.out.println("i+j" + (i+j)); >> public class Main < public static void main(String[] args) < X superOb = new X(); Y subOb = new Y(); superOb.i = 5; System.out.println("Super Class"); superOb.printi(); subOb.i = 10; subOb.j = 8; System.out.println("Sub Class"); subOb.printi(); subOb.printj(); subOb.sum(); >>

Super Class i: 5 Sub Class i: 10 j: 8 i+j18

Here subclass Y can access all members of superclass X. That is why the object of subclass Y subOb can access variable i and method printi().

Types of Inheritance

Java supports various types of inheritances.

In this type of inheritance, only one class extends another class.

Here, class X is extended by class Y. Class Y becomes subclass and class X becomes superclass.

In multiple inheritance, one class extends more than one class. Java doesn’t support multiple inheritance.

In multilevel inheritance, one class is inherited by other class. And some other class can also inherit the child class.

Class Y is subclass of class X and class Z is subclass of class Y.

In hierarchical inheritance, many classes can inherit the same class.

Here, class X, Y and Z are subclasses of same class W.

Hybrid inheritance is a combination of single and multiple inheritances. As Java doesn’t support multiple inheritances, hybrid inheritance is also not supported by Java.

Constructors Behavior for Inherited Classes

When any class hierarchy is created, constructors are called from superclass to subclass. Below example will illustrate the same.

class X < X() < System.out.println("Inside X's constructor."); >> class Y extends X < Y() < System.out.println("Inside Y's constructor."); >> class Z extends Y < Z() < System.out.println("Inside Z's constructor."); >> public class Main < public static void main(String[] args) < Z obj = new Z(); >>

Inside X’s constructor. Inside Y’s constructor. Inside Z’s constructor

Method Overriding in Java

If the subclass has the same method name and signature as the superclass, then the subclass overrides the method in the superclass. So, when the method is called from the subclass, it will always execute the code defined in the subclass method. Superclass method code will be hidden. Let us understand it more through below example:

class X < void printDetails() < System.out.println("Inside Superclass Method."); >> class Y extends X < void printDetails()< System.out.println("Inside Subclass Method"); >> public class Main < public static void main(String[] args) < Y obj = new Y(); obj.printDetails(); >>

When a subclass object is invoking the overridden method, it will always call subclass version of method. Here in this example, object of Y class will call printDetails() method defined in subclass Y only.

In Java, we have a keyword super to access superclass version of method. Let just modify above example a little and see the output.

class X < void printDetails() < System.out.println("Inside Superclass Method."); >> class Y extends X < void printDetails()< System.out.println("Inside Subclass Method"); >> public class Main < public static void main(String[] args) < Y obj = new Y(); obj.printDetails(); >>

Inside Superclass Method. Inside Subclass Method

When names and types are same for methods then it is called method overriding. But if we have other differentiator for 2 methods, then that becomes method overloading which we have already learnt in previous section.

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Method overloading and inheritance in java

Although is a subtype of , since method parameter is invariant in Java, the method doesn’t override the method . Your class has no idea, what it’s successors new methods are. is replacing a method with a different implementation.

Method overloading and inheritance in java

I have the following program but it wont compile:

public class A < public void method() < System.out.println ("bla"); >> class AX extends A < public void method(int a) < System.out.println ("Blabla"); >public static void main(String[] args) < A a2 = new AX(); a2.method(5); >> 

Only methods of class A are visible to the compiler. This is because your Object a2 is of type A according to A a2 = new AX(); . If you change this line to AX a2 = new AX(); it will work.

Maybe you confuse terms overloading and overriding .

Overloading is adding a different method with same name as existing one, that differs in input parameters and return type. It has nothing to do with inheritance. From the inheritance point of view overloading is just adding a new method. Your class A has no idea, what it’s successors new methods are.

Overriding is replacing a method with a different implementation. A knows that there exists a method, therefore you can change it in it’s successor AX .

You have two possibilities. Either define public void method(int a) in A :

public class A < public void method() < System.out.println ("bla"); >public void method(int a) < System.out.println ("Blabla from A"); >> 

because the only class, that knows about the public void method(int a) is AX .

In Java visibility determined by Object type , not by Reference type .

A a2 = new AX(); Object type A, so compiler can't find method(int a); 

Modify Inherited Methods, The subclass does not override the foo method. Instead it overrides only the protected methods that perform the series of steps ( step1(obj) , step2(obj) ,

CSAwesome: Lesson 9.3

How to Override Methods using Inheritance

Using inherited overloaded methods

public class ClassB extends ClassA < public void method(Integer d) < System.out.println("ClassB: " + d + " " + d.getClass()); >> 

ClassA a = new ClassB(); a.method(3); 

ClassA: 3 class java.lang.Integer 

My question is, why isn’t ClassB ‘s method being used? a is an instance of ClassB , and ClassB ‘s method() has an Integer parameter.

My question is, why isn’t ClassB’s method being used?

Not true. The method used is ClassB ‘s method, which it inherited from ClassA .

I think the main reason behind the confusion here is the fact that the method actually is not overridden , instead it is overloaded . Although Integer is a subtype of Number , since method parameter is invariant in Java, the method public void method(Integer d) doesn’t override the method public void method(Number n) . So, ClassB ends up having two (overloaded) methods.

Static binding is used for overloaded methods, and the method with most specific parameter type is chosen by the compiler. But in this case, why does the compiler pick public void method(Number n) instead of public void method(Integer d) . That’s because of the fact that the reference that you are using to invoke the method is of type ClassA .

ClassA a = new ClassB(); //instance is of ClassB (runtime info) a.method(3); //but the reference of type ClassA (compiletime info) 

The only method that ClassA has is public void method(Number n) , so that’s what the compiler picks up. Remember, here the expected argument type is Number , but the actual argument, the integer 3, passed is auto-boxed to the type Integer . And the reason that it works is because the method argument is covariant in Java.

Now, I think it’s clear why it prints

Your issue stems from the fact that (as quoted from the official Java Tutorials on Inheritance):

In a subclass, you can overload the methods inherited from the superclass. Such overloaded methods neither hide nor override the superclass methods—they are new methods, unique to the subclass.`

Refer to the official Java tutorials for more details: http://docs.oracle.com/javase/tutorial/java/IandI/override.html

a is of type ClassA so the methods in ClassB will not be visible to instance a unless it is declared as ClassB

will produce your expected output. Number is the supertype of Integer. So whatever you pass in will be autoboxed to appropriate subtype and the method in ClassA will be called. Try passing

a.method(3.0f) // Float a.method(3.0) // Double 

because number 3 is automatically boxed to Integer.

please see the link below: http://www.javabeat.net/articles/print.php?article_id=31

General Rule: Arguments are implicitly widened to match method parameters. It’s not legal to widen from one wrapper class to another.

Why doesn’t Java allow overriding of static methods?, Overriding depends on having an instance of a class. The point of polymorphism is that you can subclass a class and the objects implementing those

How prevent a method of super in subclass (java) [duplicate]

please note to this. I have a super class

and have a other class. inherit of class A

class B extends A < public void update(string str) > B obj = new B(); obj.update(50); // oh no, bad obj.update(""); // yes 

now, i want that when i create new from class B the update() of class A not access. -in other words i want access to the update() method of B only

You can prevent that only when you have the same method name with same arguments in super type and sub type effectively overriding it. In your case, the method the signatues are different.. they take different arguments, hence based on what you pass as argument, it calls the respective method.

Actually you can override that method and throw an exception is a solution.

class A < public void update(int num) > class B extends A < public void update(int i) < throw new UnsupprotedOperationException(); >public void update(string str) > 

Or you can implement a class with generics. This may be not working but idea works.

class A  < public void update(T num) > class B extends A  < public void update(String str) > 

Make your base class method private.

Why does java allow for overriding methods with subclass return, It’s because of the guarantees that the compiler is enforcing to ensure that client code referencing the superclass method is going to be

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