I know how to overload a method, and how to override a method. But is that possible to overload AND override a method at the same time? If yes, please give an example.
Overloading and overriding are complementary things, overloading means the same method name but different parameters, and overriding means the same method name in a subclass with the same parameters. So its not possible for overloading and overriding to happen at the same time because overloading implies different parameters.
Examples:
class A {
public void doSth() { /// }
}
class B extends A {
public void doSth() { /* method overriden */ }
public void doSth(String b) { /* method overloaded */ }
}
Cheers!
overloading and overloading are just abstractions. Overloading just means the compiler uses the name in conjunction with the types and number of parameters for addressing what function to call. In reality overloading a method is no different than naming it something different because the key the compiler uses to look up the function is a combination of name and parameter list.
Overriding is kind of the same principle except the compiler can address the overriden function with the super keyword.
So can you override an overloaded function? Yes, since the overloaded method is a completely different method in the eyes of the compiler.
It depends what you mean. A method can be an override for an overloaded method in a superclass. And you can overload a method that you are simultaneously overriding using another method.
However, you cannot have one method that is both a new overload and an override. For a method to be an override, another method with the same signature must already exist in the superclass ... and that means that this method cannot be a new override.
Yes it is possible, you can overload and override a function in the same class but you would not be able to overload a function in two different classes as it is logically not possible.
#include <bits/stdc++.h>
using namespace std;
class Base {
public:
void print()
{
cout << "Base Function" << endl;
}
void print(string name) //Overloaded Function
{
cout << "Base Function : " <<name<< endl;
}
};
class Derived : public Base {
public:
void print()
{
cout << "Derived Function" << endl;
}
void print(string name) //Overloaded Function
{
cout << "Derived Function : "<<name<< endl;
}
};
int main() {
Derived derived1, derived2;
//accesses the print function in derived class
derived1.print();
//accesses the overloaded print function in derived class
derived1.print("Gitanshu");
//accesses the print function in base class
derived2.Base::print();
//accesses the overloaded print function in base class
derived2.Base::print("Gitanshu");
return 0;
}
Output:
Derived Function
Derived Function : Gitanshu
Base Function
Base Function : Gitanshu
I answered it using C++, The same would apply in Java as well.
Related
This question already has answers here:
Is it possible to have different return types for a overloaded method?
(13 answers)
The relationship of overload and method return type in Java?
(4 answers)
Closed 6 years ago.
I am gonna put this question to have a clear idea about overloading Concept in java . As per my understanding while method resolution in overloading compiler will look for method signature that is it should have same method name and different argument types . But what if the return type is different ??
class Test{
public void m1(int i) {
System.out.println(" int arg");
}
public int m1(String s) {
System.out.println("String-arg");
return (5+10);
}
public static void main (String[] args) throws java.lang.Exception
{
Test t = new Test();
t.m1(5);
int i = t.m1("ani");
System.out.println(i);
}}
the above program is running perfectly . my doubt here is , the method m1() is it overloaded ?? it has different return type . someone please make it clear. Thanks in advance
In Java methods are identified by name and arguments' classes and amount. The return type doesn't identify the method. For this reason the following code would be illegal:
public void m1(String i) {
System.out.println(" int arg");
}
public int m1(String s) {
System.out.println("String-arg");
return (5+10);
}
If two methods of a class (whether both declared in the same class, or both inherited by a class, or one declared and one inherited) have the same name but signatures that are not override-equivalent, then the method name is said to be overloaded. (...) When a method is invoked (§15.12), the number of actual arguments (and any explicit type arguments) and the compile-time types of the arguments are used, at compile time, to determine the signature of the method that will be invoked (§15.12.2). If the method that is to be invoked is an instance method, the actual method to be invoked will be determined at run time, using dynamic method lookup (§15.12.4)
Summarizing, two methods with the same name can return different types, however it's not being taken into account when deciding which method to call. JVM first decides which method to call and later checks if the return type of that method can be assigned to the certain variable.
Example (try to avoid such constructions):
public int pingPong(int i) {
return i;
}
public String pingPong(String s) {
return s;
}
public boolean pingPong(boolean b) {
return b;
}
if we follow the Oracle definition then yes, it is a overloaded method
here the info (emphasis mine)
The Java programming language supports overloading methods, and Java
can distinguish between methods with different method signatures. This
means that methods within a class can have the same name if they have
different parameter lists (there are some qualifications to this that
will be discussed in the lesson titled "Interfaces and Inheritance").
the fact that the method return a value or not is IRRELEVANT for the overloading definition...
another thing is here why can a method somethimes return a value and sometimes no...
this will drive crazy the people using the code, but that is another question...
If I have the following code in Java:
class A {
public int add(int a , int b) {
return (a+b);
}
}
class B extends A {
public float add(float a , float b) {
return (a+b);
}
In this particular case the sub-class isn't exactly overriding the base class's add function as they have different signatures and the concept of overloading occurs only if they are in the same scope. So, is the function add(float , float) in the sub-class B treated as an entirely new function and the concept of overloading and overriding is not applicable to it? And does it use 'Static binding' or 'Dynamic Binding'?
Method add in class b is an overload of add in class a. Not an override. An override would just be a different implementation of the original add method.
In brief, yes. To override, you need to replicate the complete method signature, which includes the method name, parameters and return types. From the tutorial
An instance method in a subclass with the same signature (name, plus
the number and the type of its parameters) and return type as an
instance method in the superclass overrides the superclass's method.
You might want to consider the #Override annotation, which will trigger a compiler error if you don't successfully overrride a method.
In this particular instance, it perhaps looks like you don't need overriding so much as some solution including generics. So you could instantiate a class a<Integer> and a similar class a<Float>
In that case you are not overriding the method, since the signatures are different.
But there is overloading in class b, since you have two methods with the same name but different parameters (one if class a, and the other one in class b)
Hope it helps.
There can be a method that is not overridden but overloaded in the subclass. Here the subclass has two add() methods. The version which accepts int arguments(not overridden), and the overloaded method add() which accepts float arguments.
I think in this particular case neither overloading nor overriding occurs, because return type must be same in case overloading and overriding, so neither static binding nor dynamic binding happens in this case.
method overloading is not possible in case of different return type, because compiler can't figure that which method he need to call.
I know it's late answer but i think it's important question need to be answered for beginners.
One key point in overloading is it works in inheritance.
Next is either it's Static binding or Dynamic binding.
It is Static Binding So, why?
Static Binding
Static binding in Java occurs during Compile time.
private, final and static methods and variables uses static binding and bonded by compiler.
Static binding uses Type (class in Java) information for binding.
Dynamic Binding
Dynamic binding occurs during Runtime.
Dynamic methods bonded during runtime based upon runtime object.
Dynamic binding uses Object to resolve binding.
But the important part is here
Overloaded methods are bonded using static binding while overridden methods are bonded using dynamic binding at runtime.
Java compiler determines correct version of the overloaded method to be executed at compile time based upon the type of argument used to call the method and parameters of the overloaded methods of both these classes receive the values of arguments used in call and executes the overloaded method.
B a=new B();
a.add(4, 5);
a.add(4.0f, 5.0f);
So if you will create reference of type B then it will search for proper argument type
and for above code it will execute both methods.
A a=new B();
a.add(4, 5);
a.add(4.0f, 5.0f);
but for above code it will give compile time error for float arguments.
Hope it clears all doubts.
First things first
Is it method overriding ?
No , since to override a method you need to replicate the complete method signature as pointed out in Brian Agnew's answer and as I explain below.
Is it overloading ?
Yes , Method "add" has an overloaded implementation in Class B.
Consider the following code:
class C{
public static void main(String args[]){
B a = new B();
a.add(2 , 3);
a.add(2.0 , 3.0);
}
}
class A {
public int add(int a , int b) {
System.out.print("INT ");
return a + b;
}
}
class B extends A {
public double add(double a , double b) {
System.out.print("Double ");
return a + b;
}
}
OUTPUT : INT Double
So , the method in Class B in your code overloads the add method that it inherits from its parent
Does it use Static Binding or Dynamic Binding ?
This is what makes me conclude that OP is confused.It is static binding because it is a overloaded function. The only way to think of dynamic binding would have been in below scenario
class C{
public static void main(String args[]){
A a = new B();
a.add(2.0 , 3.0);
}
}
class A {
public int add(int a , int b) {
System.out.println("A : INT");
return a + b;
}
}
class B extends A {
public int add(int a , int b) {
System.out.println("B : INT");
return a + b;
}
public double add(double a , double b) {
System.out.println("Double");
return a + b;
}
}
Output : B : INT
Here , the parent class A has a contract that says , "I have an add behaviour for ints" . class B inherits this add behaviour and makes it more specific and at the same time also provides a new behaviour where it can add doubles.
But class A has "no knowledge of this behaviour".
So an object of class A "cannot" add doubles. To do that you need a more specific type of A object i.e. a B object.
The method add() in class A is also available to class B by inheritance, therefore the method is overloaded in class by changing the data type from int, int to float, float.
The concept of Overloading comes in play if and only if the functions are in the same scope or class.
cz if this is the case of method overloading then for same method signature or same argument type the compiler gets confuse and must give compile time error .
but in the above program in class B if you pass the same argument in same order then according to overloading it must give error but it is not happening u can check it i already have.
It is the case of inheritance where through object reference if you call any method then the compiler will check it in child class ,if its not there then it will look into parent class that the above program is all about.
hope this is helpfull.
Here's what i remembered:
In C++, if derived class defines a member function with the same name, but different signature (parameter etc) than the base class, it'll 'hide' the corresponding member function in base. e.g.
class Base {
public:
void virtual f(double x);
};
class Derived : public Base {
public:
void f(char c);
};
int main()
{
Derived* d = new Derived();
Base* b = d;
b->f(65.3); // call f(double x)
d->f(65.3); // call f(char c)
delete d;
return 0;
}
correct me if i am wrong, but i think in C++ by saying 'hide', it also means Derived class can not see 'f(double x)', or in other words, Derived do not have 'f(double x)' as its inherited member function from Base, correct?
In Java tutorial, 'hide' actually means something else (for static class method), whereas for instance method, you can overload the method inherited from base. Looking at this example: Using inherited overloaded methods
public class ClassA {
public void method(Number n) {
System.out.println("ClassA: " + n + " " + n.getClass());
}
}
public class ClassB extends ClassA {
public void method(Integer d) {
System.out.println("ClassB: " + d + " " + d.getClass());
}
}
ClassA a = new ClassB();
a.method(3);
In C++ type of thinking, i would get the same result that 'method(Number n)' from Class A is called, based on 'dynamic binding' and 'hiding' idea in C++, but:
I am still not sure how to explain it in Java. The link itself explains using 'method signature is chosen at compile time' and 'it is actually calling from Class B'; but in C++ thinking, the former is OK, but i do not think it's calling from Class B, it should be calling from Class A, right?
In Using inherited overloaded methods and Java tutorial, 'Class B' is allowed to overload function from 'Class A', and 'Class B' actually can see both 'method(Number n)' and 'method(Integer d)'. So C++ and Java are treating overloading differently, right? And why is that? Say in the C++ example, if Derived also allow overloading, 'd->f(65.3)' would call 'f(double x)', not 'f(char c)'.
Thanks,
In C++ every thing is static for non-virtual functions, so for non-virtual functions you don't have dynamic-binding. And hiding do not remove function from inheritance. It's something like this:
When you define method M as: void M(int) then compiler implement a function internally name Base::M as void Base::M( Base* this, int ). Now this function is implemented somewhere in code and can't removed and you can call it as long as you be able to provide a this (actually you can call it without even having this). So in Child I can call Base::M(0); and C++ convert this from Child* to Base* and call M. When you define a function with its name correspond to the name of base class you tell the compiler that I prefer to use that name in my class for a new method or property! But you don't remove any thing and you can use using to bring old definition of M to Child:
struct Base {
void f( int ) {}
};
struct Child : Base {
void f( char* ) {}
using Base::f; // Bring Base::f to this class, so I have f(char*), f(int)
};
And beside that you can call f(int) without even use of using.
// In the Child class
void test() {
Base::f('c'); // Call Base::f(char)
}
// Outside of class
Child c;
((Base*)&c)->f('1');
This is not how I would expect C++ to behave. Rather than hiding the base class method, I would expect the subclass to simply overload the method because the parameters are different. Thus, the compiler will know which method to use when you call b->f() because there's only one available, but it has to work out which one to use when you call d->f() based in your parameter type. If it can't then at compile time you will be forced to use a cast. With your example it should recognise the decimal point and use double instead if char.
Guys I know this question is silly but just to make sure:
Having in my class method:
boolean equals(Document d)
{
//do something
}
I'm overloading this method nor overriding right? I know that this or similiar question will be on upcoming egzam and would be stupid to not get points for such a simple mistake;
Based on the code provided, we can't tell for sure whether you're overloading or overriding it.
You are most likely overloading the equals(Object o) method.
class A {
void method() {..}
}
class B extends A {
// this is overriding
void method() {..}
}
And
// this is overloading
class A {
void method(boolean b) {..}
void method(String arg) {..}
void method(int arg) {..}
}
P.S. you are using a bracket convention that is not widely accepted on the java world. In Java it is more common to place opening the curly bracket on the same.
You are not even overloading, since the other method is called equals. But if you add that s, you will be overloading equals. Although, to be precise, we talk about overloading if two (or more) methods with the same name but different signature are defined in the same class. In your case, it is trickier, since your equals with its different signature partly hides the original equals. Which is usually a bad thing, because this almost always leads to hard to understand behaviour, thus subtle bugs. Whenever someone calls equals on an instance of your class, depending on the parameter type the call may go to a different implementation of the method.
class Document {
public boolean equals(Document d)
{
//do something
}
}
Document doc1 = new Document();
Document doc2 = new Document();
String string = new String();
doc1.equals(doc2); // calls Document.equals
doc1.equals(string); // calls Object.equals!
You would be overriding Object.equals if you defined your method with the exact same signature as the original, i.e.
public boolean equals(Object obj) ...
In this case, both of the above calls to equals correctly execute Document.equals.
From the code you posted it could be either. If equal is defined in a superclass with the same parameter declarations then you are overriding it. If there is already a method called equal, but with different parameter types, you are overloading it.
On a related note, if you are using Java 5 and above and your intent is to override then it is highly recommended to use the #Override annotation before the method definition to indicate your intention. The wrong usage of this annotation (i.e. when you want to override and are not doing so) would flag a compile error.
As of Java 6 you can use the #Override annotation while defining methods that are
declared in an interface the class in implementing.
Overloading: same method name, same parameter list, different classes
Overriding: same method name, different parameter list, same or different classes.
Class A {
bool Equals(Document d) {...}
bool Equals(A a) {...} // overloaded method
}
Class B extends A {
bool Equals(Document d) {...} // overridden method
bool Equals(B b) {...} // overloaded method
}
One thing to note, the return type does not matter, it's the name of the method and the parameter list that make all the difference.
I've run into Java code similar to the following:
public interface BaseArg {
}
public class DerivedArg implements BaseArg {
}
public abstract class Base <A extends BaseArg> {
A arg;
void doIt() {
printArg(arg);
}
void printArg(A a) {
System.out.println("Base: " + a);
}
}
public class Derived extends Base<DerivedArg> {
void printArg(DerivedArg a) {
System.out.println("Derived: " + a);
}
public static void main(String[] args) {
Derived d = new Derived();
d.arg = new DerivedArg();
d.doIt();
}
}
(feel free to split it into files and run it).
This code ends up invoking the Derived printArg. I realize it's the only logical thing to do. However, if I perform "erasure" on the generic Base manually, replacing all occurrences of A with BaseArg, the overriding breaks down. I now get the Base's version of printIt.
Seems like "erasure" is not total - somehow printArg(A a) is not the same as printArg(BaseArg a). I can't find any basis for this in the language spec...
What am I missing in the language spec? It's not really important, but it bugs me :) .
Please note that the derived method is invoked. The question is why, considering their erased signatures are not override-equivalent.
When compiling class Derived, the compiler actually emits two methods: The method printArg(DerivedArg), and a synthetic method printArg(BaseArg), which overrides the superclass method in terms even a virtual machine ignorant of type parameters can understand, and delegates to printArg(DerivedArg). You can verify this by throwing an exception in printArt(DerivedArg), while calling it on a reference of type Base, and examining the stack trace:
Exception in thread "main" java.lang.RuntimeException
at Derived.printArg(Test.java:28)
at Derived.printArg(Test.java:1) << synthetic
at Base.doIt(Test.java:14)
at Test.main(Test.java:39)
As for finding this in the Java Language Specification, I first missed it as well, as it is not, as one might expect, specified where overriding or the subsignature relation is discussed, but in "Members and Constructors of Parameterized Types" (§4.5.2), which reveals that formal type parameters of the superclass are syntactically replaced by the actual type parameter in the subclass prior to checking for override equivalence.
That is, override equivalence is not affected by erasure, contrary to popular assumption.
If you do "manual" type erasure, you define the arg instance in BaseArg as type "BaseArg", not type "DerivedArg", so that's resolved to Base's "doIt(BaseArg)" method rather than Derived's "doIt(DerivedArg)" method. If you then alter Derived's method signature to
void printArg( BaseArg a )
from
void printArg(DerivedArg a)
it will print "Derived: arg" as expected.
I believe the behaviour that you encountered is due to the overloading method resolution.
See Java Lang Spec on overloading: link text
And also this wonderful resource on Java Generic regarding the topic.
The printArg in Derived does not override the printArg in Base. In order for it to override, by JLS 8.4.8.1, the overriding method's signature must be a "subsignature" of the overridden method's. And then by JLS 8.4.2, a subsignature must either have the same argument types (which yours doesn't), or its erasure must be the same (which is also not true).
First of all, you can compile the source code in a single file if you get rid of the "public" declarations for all of the classes/interfaces except "Derived".
Second, go ahead and do the type erasure by hand. Here's what I got when I did it:
interface BaseArg {}
class DerivedArg implements BaseArg {}
abstract class Base {
BaseArg arg;
void doIt() {
printArg(arg);
}
void printArg(BaseArg a) {
System.out.println("Base: " + a);
}
}
public class Derived extends Base {
void printArg(BaseArg a) {
System.out.println("Derived: " + a);
}
public static void main(String[] args) {
Derived d = new Derived();
d.arg = new DerivedArg();
d.doIt();
}
}
In the generic version of the code, it may look like methods Derived.printArg and Base.printArg have different signatures. However, if that were the case, then Derived.printArg could never be invoked by doIt. The type-erased version of the code makes it clear that Derived.printArg overrides Base.printArg, so doIt polymorphically calls the right method.
How is printArg in Base defined after your manual erasure ?
void printArg(BaseArg a) {
so, printArg(Derived a) does NOT override it and will not be called.
EDIT:
if you use the Override annotation in Derived, you'll get an error doing the manual erasure.