Suppose i have two methods in a class say
public void eat(int i,String s) and
public void eat(String s, int i)
then what is it like. Overloading or overriding?
Overloading means two methods or more with the same name but with different parameters just like your example.While Overriding you implement a method from an interface or abstract class so the method in the super class has an implementation and in the subclass has a different one, Still they have the same method name and parameters.
That would be method overloading, as it meets the conditions for method overloading:
Must have different argument lists
May have different return types, if
argument lists are also different
May have different access modifiers
May throw different exceptions
Also overriding can happen only when inheritance is involved. Since both of your methods are in the same class it cannot be overriding.
This is overloading. Overriding is used in inheritance when you give different implementation to the same method signature.
That's overloading. In brief:
Overriding = replacing
Overloading = give more options
Method Overloading simply means providing two separate methods in a class with the same name but different arguments while method return type may or may not be different which allows us to reuse the same method name.
Method Overriding means defining a method in the child class which is already defined in the parent class with same method signature i.e same name, arguments and return type.
Difference Between Method Overloading and Method Overriding
For more details, you can read Everything About Method Overloading Vs Method Overriding.
Kind of rules about overloading and overriding:
Constructors can be overloaded but not overridden.
Abstract methods must be overridden by the first concrete sub-class.
The overriding method:
must have same argument list,
must have same return type (it can also be a subclass of parent's class' return type,
must not have a more restrictive access modifier,
may have a less restrictive access modifier,
must not throw new or broader checked exceptions,
may throw fewer or narrower checked exceptions, or any unchecked exception.
final methods cannot be overridden.
Only inherited methods may be overridden, and remember that private methods are not inherited.
In a subclass use: super.overriddenMethodName() to call superclass' overridden method.
Overloaded methods:
must have different argument lists,
may have different return types, if argument lists are also different,
may have different access modifiers,
may throw different exceptions.
Methods from a superclass can be overloaded in a subclass.
Polymorphism applies to overriding but not to overloading.
Object type (not the reference variable's type) determines which overridden method is used at runtime.
Reference type determines which overloaded method will be used at compile time.
*Taken from Sun Certified Programmer for Java 6 Study Guide by Kathy Seirra, Bert Bates.
Overloading : In Java Overloading is a Process where a class Contains multiple implementation of methods or constructor by differentiating there no of arguments or types of arguments.
public class TestDemo
{
public void disp(String c)
{
System.out.println(c);
}
public void disp(String c, int n)
{
System.out.println(c+" "+n);
} }
class Sample
{
public static void main(String[] args)
{
// TODO Auto-generated method stub
TestDemo obj=new TestDemo();
obj.disp("a");
obj.disp("a",2);
}}
Overriding : Here Methods can be over riding. It must be happen within two class. One is parent & another is child. In this parent class’s method with same name and signature.
public class TestDemo
{
public void eat()
{
System.out.println("Human is Eating");
}
}
class Overriding extends TestDemo
{
public void eat()
{
System.out.println("Human is Eating");
}
public static void main(String[] args)
{
Overriding obj=new Overriding();
obj.eat();
}
}
for more knowledge click on this link : https://arzatechs.com/overloading-and-overriding-in-java/
OVERLOADING:
Two or more methods with same name but different parameters in same class. The same as your question situation.
public void eat(int i,String s)
public void eat(String s, int i)
So, in your case it is method Overloading.
OVERRIDING:
Overriding means having two methods with the same method name and parameters (i.e., method signature). One of the methods is in the parent class and the other is in the child class.
ex.
class Animal{
void eat(){System.out.println("eating...");}
}
class Dog extends Animal{
void eat(){System.out.println("eating bread...");}
}
In this specific case, it's overloading.
Let's differentiate both terms a bit dipper:
Overloading (same function name but different signature):
Two or more methods having the same name with different arguments in
same class.
Overloading is used when you want to extend class functionality.
Overloading is is a compile time polymorphism.
Overriding (same function with the same signature):
Two or more methods having the same method name and same arguments in parent class and child class.
Overriding is used when you want to reuse the existing functionality.
Overriding is a run time polymorphism.
Visit to learn more about overloading and overriding.
Property ------------------- OverLoading -------------------- Overriding
Method Names -------------->must be Same--------------------> must be same
Arg Types------------------>must be Different(atleast arg)---->must be same(Including Order)
Method Signature----------->must be Different(atleast arg)---->must be same(Including Order)
Return Type---------------->No restriction------------------->No restriction
Private,Static,Final------->No Restriction-------------------->Must be Same
Access Modifyer------------>No Restriction--------------------Same
try/Catch----------------->No Restriction--------------------Exception Is thrown
Method Resolution-------->Compiler time (Reference)--------(JVM) Run Time Poymorphism
Related
So, I was studying from Head First Java and there I read -
An overloaded method is just a different method that happens to have the same method name, It has nothing to do with inheritance and polymorphism.
However, I have another famous book named as Java - The Complete Reference by Herbert Schildt and in that book, I read:
Method overloading is one of the ways that Java supports polymorphism. Method overloading supports polymorphism because it is one way that Java implements the “one interface, multiple methods” paradigm.
Which one of them is correct?
To answer your question in the title, no, you cannot use method overloading to describe polymorphism.
Method Overloading is when a class has two or more methods by the same name, the only difference is its parameters. This is a completely difference concept than polymporhism.
Polymorphism is the ability of an object to take on many forms. This occurs when a parent class reference is used to refer to a child class object.
For example, if I was tasked with creating some type of application for a zoo and I wanted to keep track of the animals using classes. I would make a class called "Animal" and additional classes for each animal, "Lion", "Eagle", "Dolphin". After declaring the animal classes as subclasses, I can then use the concept of polymorphism to create Lions, Eagles, and Dolphins which are of type "Animal".
Can method overloading be considered a way of describing polymorphism
in Java?
No these are two different concepts.
I think that you incorrectly understood what the author means.
I stressed in bold the important part :
Method overloading is one of the ways that Java supports polymorphism.
Method overloading supports polymorphism because it is one way that
Java implements the “one interface, multiple methods” paradigm.
It simply means that overloading and polymorphism are not incompatible.
For example, you may have an interface that defines an overloaded method :
public interface Doable{
void doX(String s);
void doX(Integer s)
}
Suppose MyDoable an implementation of it and MyChildDoable a subclass of MyDoable.
When you write :
MyDoable myDoable = new MyChildDoable();
myDoable.doX("a")
the compiler will bound the invoked method to void doX(String s); (overloading) but at runtime, the method of the instance (MyChildDoable) will be used (polymorshism).
Polymorphism is not of one type, there are different types of polymorphism. Please refer this wonderful link - Java Polymorphism to understand it better.
Overloading supports polymorphism, the polymorphic behavior invocation is based on type arguments passed and it takes place in an ad-hoc way.
So, there is this idea of compile time polymorphism and runtime polymorphism. The idea of polymorphism implies same interface but different implementations.
The way "compile time" polymorphism works is through function overloading i.e. you have the same function name but with different arrity. I am guessing this is what the second second book is referring to as "polymorphism supported through method overloading". However, strictly speaking the interface also includes the arrity, which means method overloading might not constitute polymorphism.
Runtime polymorphism is where the actual type of a given object is what dictates what implementation should be used. This is where the interface / base class and derived class comes into picture. This would be the more accurate definition of polymorphism - at least from an object orientation point of view.
Overriding - this is about polymorphism.
public class Base {
public void print() {
System.out.println("base");
}
}
public class Extend extends Base {
#Override
public void print() {
System.out.println("Extend");
}
}
Let's look at the output
new Base().print();
will print "Base"
new Extend().print();
will print "Extend".
Overloading - this is about method overload in one class with different parameters:
public class ABC {
// method 1
void method(int a){
System.out.println("method 1");
}
// method 2
void method(int a, int b){
System.out.println("method 2");
}
}
Let's look at the output
ABC abc = new ABC();
abc.method(5);
will invoke method(int) and print "method 1"
abc.method(1,2)
will invoke method(int, int) and print "method 2"
Overriding - this is runtime feature;
Overloading - this compile time feature;
all overloading method (with one name) must be same return type or void.
Below is the code which I am trying to analyze.
class Acc{
public void aMethod(List<Integer> A){
System.out.println(A.toString());
}
}
class Back extends Acc{
public void aMethod(String A){
System.out.println(A);
}
}
Here if I invoke it as
Acc a = new Back();
a.aMethod(list);
But upon debugging the method of parent class is being called.But it should dynamically call method of Back as per my understanding.Please help me understand of this behaviour.Now again If I keep parent class argument as List<Integer> and child one as List<String> then it gives a name clash thing.Why not in this case?
Parent class method public void aMethod(List<Integer> A) and child class method is public void aMethod(String A).
Both have different parameters, so it is not Method Overriding. However in child class it can be thought of as Method Overloading.
When you call the method, you are passing the list parameter and there is only only method that matches this signature and that is of parent class. So the method of parent class gets called.
Method Overriding
For method overriding methods in base class and child class must have same signature i.e. same name and same arguments. However method in child class can have higher scope access specifier than base class's method.
In simpler words, if method in base class is default then method in child class can be public.
Here are some rules for method overriding:
The argument list should be exactly the same as that of the overridden method.
The return type should be the same or a subtype of the return type declared in the original overridden method in the superclass.
The access level cannot be more restrictive than the overridden method's access level. For example: if the superclass method is declared public then the overridding method in the sub class cannot be either private or protected.
Instance methods can be overridden only if they are inherited by the subclass.
A method declared final cannot be overridden.
A method declared static cannot be overridden but can be re-declared.
If a method cannot be inherited, then it cannot be overridden.
A subclass within the same package as the instance's superclass can override any superclass method that is not declared private or final.
A subclass in a different package can only override the non-final methods declared public or protected.
An overriding method can throw any uncheck exceptions, regardless of whether the overridden method throws exceptions or not. However the overriding method should not throw checked exceptions that are new or broader than the ones declared by the overridden method. The overriding method can throw narrower or fewer exceptions than the overridden method.
If you check this and then code written, it would be clear where its wrong.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Do interfaces inherit from Object class in java
package inheritance;
class A{
public String display(){
return "This is A!";
}
}
interface Workable{
public String work();
}
class B extends A implements Workable{
public String work(){
return "B is working!";
}
}
public class TestInterfaceObject{
public static void main(String... args){
B obj=new B();
Workable w=obj;
//System.out.println(w.work());
//invoking work method on Workable type reference
System.out.println(w.display());
//invoking display method on Workable type reference
//System.out.println(w.hashCode());
// invoking Object's hashCode method on Workable type reference
}
}
As we know that methods which can be invoked depend upon the type of the reference variable on which we are going to invoke. Here, in the code, work() method was invoked on "w" reference (which is Workable type) so method invoking will compile successfully. Then, display() method is invoked on "w" which yields a compilation error which says display method was not found, quite obvious as Workable doesn't know about it. Then we try to invoke the Object class's method i.e. hashCode() which yields a successful compilation and execution. How is it possible? Any logical explanation?
The intuitive answer is that regardless of what interface you refer to, the object implementing the interface must be a subclass of Object.
Section 9.2 of the JLS specifically defines this behaviour: http://docs.oracle.com/javase/specs/jls/se7/html/jls-9.html#jls-9.2
If an interface has no direct superinterfaces, then the interface implicitly declares a public abstract member method m with signature s, return type r, and throws clause t corresponding to each public instance method m with signature s, return type r, and throws clause t declared in Object, unless a method with the same signature, same return type, and a compatible throws clause is explicitly declared by the interface.
i.e. all interfaces are assumed to contain method signatures corresponding to methods in the Object class.
I think what's happening here is that even though w is known only to be Workable, all objects must derive from Object, so no matter what class w eventually is, it must have the Object methods.
The reason w.display() doesnt work is as you have save the reference as your interface type. The compiler only sees the methods exposed by the interface. If you were to call ((B)w).display() this would work. You are able to call hashCode() as the compiler is smart enough to know that interfaces are inherited by Objects and all object's superclass is Object
If I had a method signature
public void myMethod(SuperClass s){
}
and SuperClass has three subclasses, is there any way within myMethod I can get the class name of the subclass which was passed in?
Not sure if it's important, but SuperClass is abstract.
is there any way within myMethod I can get the class name of the subclass which was passed in?
Yes, by using the getClass method:
public void myMethod(SuperClass s){
System.out.println(s.getClass());
}
Remark 1:
This does however sound to me like a Bad Design™.
Whatever you want to do in myMethod consider having a method for it in SuperClass (provide a meaningful default implementation, or make it abstract to force subclasses to implement the method) and call this method on s in your method:
public void myMethod(SuperClass s){
s.abstractMethod();
}
Remark 2:
If myMethods logic is seemingly unrelated to the purpose of the SuperClass and you don't want to put the myMethod code inside this class, consider implementing the visitor pattern instead.
We always say that method overloading is static polymorphism and overriding is runtime polymorphism. What exactly do we mean by static here? Is the call to a method resolved on compiling the code? So whats the difference between normal method call and calling a final method? Which one is linked at compile time?
Method overloading means making multiple versions of a function based on the inputs. For example:
public Double doSomething(Double x) { ... }
public Object doSomething(Object y) { ... }
The choice of which method to call is made at compile time. For example:
Double obj1 = new Double();
doSomething(obj1); // calls the Double version
Object obj2 = new Object();
doSomething(obj2); // calls the Object version
Object obj3 = new Double();
doSomething(obj3); // calls the Object version because the compilers see the
// type as Object
// This makes more sense when you consider something like
public void myMethod(Object o) {
doSomething(o);
}
myMethod(new Double(5));
// inside the call to myMethod, it sees only that it has an Object
// it can't tell that it's a Double at compile time
Method Overriding means defining a new version of the method by a subclass of the original
class Parent {
public void myMethod() { ... }
}
class Child extends Parent {
#Override
public void myMethod() { ... }
}
Parent p = new Parent();
p.myMethod(); // calls Parent's myMethod
Child c = new Child();
c.myMethod(); // calls Child's myMethod
Parent pc = new Child();
pc.myMethod(); // call's Child's myMethod because the type is checked at runtime
// rather than compile time
I hope that helps
Your are right - calls to overloaded methods are realized at compile time. That's why it is static.
Calls to overridden methods are realized at run-time, based on the type on which the method is invoked.
On virtual methods wikipedia says:
In Java, all non-static methods are by default "virtual functions." Only methods marked with the keyword final are non-virtual.
final methods cannot be overridden, so they are realized statically.
Imagine the method:
public String analyze(Interface i) {
i.analyze();
return i.getAnalysisDetails();
}
The compiler can't overload this method for all implementations of Interface that can possibly be passed to it.
I don't think you can call overloading any sort of polymorphism. Overloaded methods are linked at compile time, which kind of precludes calling it polymorphism.
Polymorphism refers to the dynamic binding of a method to its call when you use a base class reference for a derived class object. Overriding methods is how you implement this polymorphic behaviour.
i agree with rachel, because in K&B book it is directly mentioned that overloading does not belong to polymorphism in chapter 2(object orientation). But in lots of places i found that overloading means static polymorphism because it is compile time and overriding means dynamic polymorphism because it s run time.
But one interesting thing is in a C++ book (Object-Oriented Programming in C++ - Robert Lafore) it is also directly mentioned that overloading means static polymorphism.
But one more thing is there java and c++ both are two different programing languages and they have different object manipulation techniques so may be polymorphism differs in c++ and java ?
Method Overloading simply means providing two separate methods in a class with the same name but different arguments while method return type may or may not be different which allows us to reuse the same method name.
But both methods are different hence can be resolved by compiler at compile time that's is why it is also known as Compile Time Polymorphism or Static Polymorphism
Method Overriding means defining a method in the child class which is already defined in the parent class with same method signature i.e same name, arguments and return type.
Mammal mammal = new Cat();
System.out.println(mammal.speak());
At the line mammal.speak() compiler says the speak() method of reference type Mammal is getting called, so for compiler this call is Mammal.speak().
But at the execution time JVM knows clearly that mammal reference is holding the reference of object of Cat, so for JVM this call is Cat.speak().
Because method call is getting resolved at runtime by JVM that's why it is also known as Runtime Polymorphism and Dynamic Method Dispatch.
Difference Between Method Overloading and Method Overriding
For more details, you can read Everything About Method Overloading Vs Method Overriding.
Simple Definition - Method overloading deals with the notion of having two or more methods(functions) in the same class with the same name but different arguments.
While Method overriding means having two methods with the same arguments, but different implementation. One of them would exist in the Parent class (Base Class) while another will be in the derived class(Child Class).#Override annotation is required for this.
Check this :
Click here for a detailed example
Property Over-loading Overriding
Method Names -------------->must be Same----------------must be same
Arg Types------------------>must be Different(at least arg)
Method Signature
Return Type
Private,Static,Final
Access Modifier
try/Catch
Method Resolution
First, I want to discuss Run-time/Dynamic polymorphism and Compile-time/static polymorphism.
Compile-time/static polymorphism:- as its name suggests that it bind the function call to its appropriate Function at compile time. That means the compiler exactly know which function call associated to which function. Function overloading is an example of compile time polymorphism.
Run-time/Dynamic polymorphism:-In this type of polymorphism compiler don't know which functions call associates to which function until the run of the program. Eg. function overriding.
NOW, what are the function overriding and function overloading???
Function Overloading:- same function name but different function signature/parameter.
eg. Area(no. of parameter)
{ -------------
----------------
return area;}
area of square requires only one parameter
area of rectangle requires two parameters(Length and breadth)
function overriding:- alter the work of a function which is present in both the Superclass and Child class.
eg. name() in superclass prints "hello Rahul" but after overring in child class it prints "hello Akshit"
Tried to cover all differences
Overloading Overriding
Method Name Must be same Must be same
Argument Types Must be same Must be different
Return Type No restriction Must be same till 1.4V
but after 1.4V
co- variants
were introduced
private/static/final Can be overloaded Cannot be overridden
Access Modifiers No restriction Cannot reduce the scope
Throws keyword No restriction If child class method
throws a checked
exception the parent
class method must throw
the same or the
parent exception
Method Resolution Taken care by compiler Taken care by JVM based
based on reference types on run-time object
Known as Compile-Time Polymorphism, RunTime Polymorphism,
Static Polymorphism, or dynamic polymorphism,
early binding late binding.