I know that in case of dynamic binding only those method which are present in current class can be called. If child override parent method then the child method is executed otherwise parents method will be executed...
But in case of interfaces what is happening I don't know... Here's an example of what I mean:
interface TestInterface {
public void show();
}
class Test implements TestInterface {
public void show() {
System.out.println("Hello in show");
}
public String toString() {
System.out.println("Hello in To String");
return "";
}
public static void main(String[] args) {
TestInterface obj = new Test();
obj.show();
obj.toString(); // why it run vilate dynamic binding rule..
}
}
This is because interfaces implicitly include all public methods declared in Object.
This is specified in the JLS, section 9.2 Interface Members.
9.2 Interface Members
[...]
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.
[...]
In case of interfaces in java :-"All interfaces get all public and abstract method of Object class"
Because in case of interfaces its implicitly include all public methods declared in Object
Your code does not compile. I changed your code to:
interface TestInterface {
public void show();
}
class Test implements TestInterface {
#Override
public void show() {
System.out.println("Hello in show");
}
#Override
public String toString() {
return "Hello in To String";
}
public static void main(String[] args) {
TestInterface obj = new Test();
obj.show();
System.out.println(obj.toString());
}
}
The result is:
Hello in show
Hello in To String
Related
This question already has answers here:
Java superclass calls subclass method
(5 answers)
Closed 1 year ago.
Why does method() call the overridden subclass method2 instead of method2 in the BaseClass?
public class BaseClass {
public void method(){
System.out.println("method() called");
method2();
}
public void method2(){
System.out.println("method2() called");
}
}
public class ChildClass extends BaseClass {
public void method2(){
System.out.println("method2() from BaseClass");
}
}
public class Main {
public static void main(String[] args) {
ChildClass obj = new ChildClass();
obj.method();
}
}
This is the concept of Runtime polymorphism (Dynamic Method Dispatch). Because you are assigning the object (instance) of ChildClass to obj reference variable, it will call the method of child class.
Always the method of the class whose instance is created gets called first. If that method is not present in that particular child class, then the parent's inherited method gets called.
If you come from the C++ corner:
all instance methods (non-static) in Java are virtual.
All class methods (static) are NOT.
This is why your case happens.
This is also, why the Java compiler will complain (warn) that if you access a static method via an object, that you should call via the distinct class, because calls to the "static method of a object" could be ambiguous, because it could be two static methods with the same signature that get called.
Extending your Example:
package stackoverflow.staticcalls;
public class BaseClass {
public void method() {
System.out.println("method() called");
method2();
}
public void method2() {
System.out.println("method2() called");
}
static public void sayHello() {
System.out.println("BaseClass.sayHello()");
}
}
and
package stackoverflow.staticcalls;
public class ChildClass extends BaseClass {
public void method2() { // compiler warning: The method method2() of type ChildClass should be tagged with #Override since it actually overrides a superclass method
System.out.println("method2() from BaseClass");
}
public void originalCallToBaseMethod2() {
super.method2(); // will run BaseClass.method2()
}
static public void sayHello() {
System.out.println("ChildClass.sayHello()");
}
}
and
package stackoverflow.staticcalls;
public class Main {
public static void main(final String[] args) {
final ChildClass obj = new ChildClass();
System.out.println("\nCalling obj.method(); ...");
obj.method();
System.out.println("\nCalling obj.sayHello(); ...");
obj.sayHello(); // compiler warning: The static method sayHello() from the type ChildClass should be accessed in a static way
System.out.println("\nCalling ChildClass.sayHello(); ...");
ChildClass.sayHello(); // the proper call
System.out.println("\nCalling BaseClass.sayHello(); ...");
BaseClass.sayHello(); // but you can also explpicitly call the other method
System.out.println("\nCalling obj.originalCallToBaseMethod2(); ...");
obj.originalCallToBaseMethod2(); //
}
}
Here you see the examples to what I said.
Note: In the last call in Main.main() we still can call BaseClass.method2(), but not directly. We have to be within ChildClass to do that, and it's done via the super keyword/reference.
A little off-topic note, to complete addressing patterns:
If you're inside an inner class and need to call to a overshadowed name in the outer class, you can use Outer.this.method():
package stackoverflow.staticcalls;
import stackoverflow.staticcalls.OuterInner.Outer.Inner;
public class OuterInner {
class Outer {
void method() {
System.out.println("OuterInner.Outer.method()");
}
class Inner {
void method() {
System.out.println("OuterInner.Outer.Inner.method()");
}
void callOuter() {
Outer.this.method();
}
}
}
public static void main(final String[] args) {
final OuterInner oi = new OuterInner();
final Outer outer = oi.new Outer();
final Inner inner = outer.new Inner();
System.out.println("\nCalling inner.method(); ...");
inner.method();
System.out.println("\nCalling inner.callOuter(); ...");
inner.callOuter();
}
}
Consider the following:
//Fooable.java
public interface Fooable {
public default void foo() {
System.out.println("Fooable::foo");
}
//Lots of other non-default methods...
}
//MyFooable.java
public class MyFooable implements Fooable {
#Override
public void foo() {
System.out.println("MyFooable::foo");
}
//implements other methods in Fooable...
}
//MyAdvancedFooable.java
public class MyAdvancedFooable extends MyFooable {
#Override
public void foo() {
Fooable.super.foo();
System.out.println("MyAdvancedFooable::foo");
}
public static void main(String[] args) {
new MyAdvancedFooable().foo();
}
}
As you can see, I want to call foo() in Fooable from MyAdvancedFooable (a subclass of MyFooable). However, when I try to compile, I get the following error:
MyAdvancedFooable.java:4: error: not an enclosing class: Fooable
Fooable.super.foo();
if I try MyAdvancedFooable extends MyFooable implements Fooable I get the following:
MyAdvancedFooable.java:4: error: bad type qualifier Fooable in default super call
Fooable.super.foo();
method foo() is overridden in MyFooable
How can I resolve this problem without having to create a new anonymous implementation of Fooable?
You can only call a method one level up so you would need
Fooable.super.foo();
in MyFooable, while just calling super.foo() in MyAdvancedFooable
You need to use just super.foo() or this.super.foo() as it is the parent of the object and not of the class as implied by Fooable.super.foo().
I have a class called "Design", and I am writing the following code to extend the class to include a new function called sayHello(). However it doesnt seem to be working. Am I only allowed to ovveride existing functions in that way?
Design design1 = new Design() {
public void sayHello() {
System.out.println("hello");
}
};
design1.sayHello(); // this gives an error "function not found"
So that is logical.
You are overriding a super class and assigning it to the instance variable of super class.
As the instance variable is of super class type, you can access only those methods which are available in super class.
You can do that if your "Design" class is an interface. You could create an anonymous inner class.
Example:
public interface Design{
public void sayHello();
}
public class Test
{
public static void main(String args[])
{
Design d = new Design(){
#Override
public void sayHello(){
return "Hello World!";
}
};
System.out.println(d.sayHello());
}
}
Java 8 introduces default methods to provide the ability to extend interfaces without the need to modify existing implementations.
I wonder if it's possible to explicitly invoke the default implementation of a method when that method has been overridden or is not available because of conflicting default implementations in different interfaces.
interface A {
default void foo() {
System.out.println("A.foo");
}
}
class B implements A {
#Override
public void foo() {
System.out.println("B.foo");
}
public void afoo() {
// how to invoke A.foo() here?
}
}
Considering the code above, how would you call A.foo() from a method of class B?
As per this article you access default method in interface A using
A.super.foo();
This could be used as follows (assuming interfaces A and C both have default methods foo())
public class ChildClass implements A, C {
#Override
public void foo() {
//you could completely override the default implementations
doSomethingElse();
//or manage conflicts between the same method foo() in both A and C
A.super.foo();
}
public void bah() {
A.super.foo(); //original foo() from A accessed
C.super.foo(); //original foo() from C accessed
}
}
A and C can both have .foo() methods and the specific default implementation can be chosen or you can use one (or both) as part of your new foo() method. You can also use the same syntax to access the default versions in other methods in your implementing class.
Formal description of the method invocation syntax can be found in the chapter 15 of the JLS.
This answer is written mainly for users who are coming from question 45047550 which is closed.
Java 8 interfaces introduce some aspects of multiple inheritance. Default methods have an implemented function body. To call a method from the super class you can use the keyword super, but if you want to make this with a super interface it's required to name it explicitly.
class ParentClass {
public void hello() {
System.out.println("Hello ParentClass!");
}
}
interface InterfaceFoo {
public default void hello() {
System.out.println("Hello InterfaceFoo!");
}
}
interface InterfaceBar {
public default void hello() {
System.out.println("Hello InterfaceBar!");
}
}
public class Example extends ParentClass implements InterfaceFoo, InterfaceBar {
public void hello() {
super.hello(); // (note: ParentClass.super could not be used)
InterfaceFoo.super.hello();
InterfaceBar.super.hello();
}
public static void main(String[] args) {
new Example().hello();
}
}
Output:
Hello ParentClass!
Hello InterfaceFoo!
Hello InterfaceBar!
The code below should work.
public class B implements A {
#Override
public void foo() {
System.out.println("B.foo");
}
void aFoo() {
A.super.foo();
}
public static void main(String[] args) {
B b = new B();
b.foo();
b.aFoo();
}
}
interface A {
default void foo() {
System.out.println("A.foo");
}
}
Output:
B.foo
A.foo
You don't need to override the default method of an interface. Just call it like the following:
public class B implements A {
#Override
public void foo() {
System.out.println("B.foo");
}
public void afoo() {
A.super.foo();
}
public static void main(String[] args) {
B b=new B();
b.afoo();
}
}
Output:
A.foo
It depends on your choice whether you want to override the default method of an interface or not. Because default are similar to instance method of a class which can be directly called upon the implementing class object. (In short default method of an interface is inherited by implementing class)
Consider the following example:
interface I{
default void print(){
System.out.println("Interface");
}
}
abstract class Abs{
public void print(){
System.out.println("Abstract");
}
}
public class Test extends Abs implements I{
public static void main(String[] args) throws ExecutionException, InterruptedException
{
Test t = new Test();
t.print();// calls Abstract's print method and How to call interface's defaut method?
}
}
Can private methods be overridden in Java?
If no, then how does the following code work?
class Base{
private void func(){
System.out.println("In Base Class func method !!");
};
}
class Derived extends Base{
public void func(){ // Is this a Method Overriding..????
System.out.println("In Derived Class func method");
}
}
class InheritDemo{
public static void main(String [] args){
Derived d = new Derived();
d.func();
}
}
No, you are not overriding it. You can check by trying to mark it with #Override, or by trying to make a call to super.func();. Both won't work; they throw compiler errors.
Furthermore, check this out:
class Base {
private void func(){
System.out.println("In base func method");
};
public void func2() {
System.out.println("func2");
func();
}
}
class Derived extends Base {
public void func(){ // Is this an overriding method?
System.out.println("In Derived Class func method");
}
}
class InheritDemo {
public static void main(String [] args) {
Derived D = new Derived();
D.func2();
}
}
It will print:
func2
In base func method
When you change func() in Base to public, then it will be an override, and the output will change to:
func2
In Derived Class func method
No, a private method cannot be overridden because the subclass doesn't inherit its parent's private members. You have declared a new method for your subclass that has no relation to the superclass method. One way to look at it is to ask yourself whether it would be legal to write super.func() in the Derived class. There is no way an overriding method would be banned from accessing the method it is overriding, but this would precisely be the case here.
No, it is not. You can mark an override just to make sure like this:
#Override
public void func(){
System.out.println("In Derived Class func method");
}
And in this case it would be a compiler error.
You are not overriding. You cannot override private members, you are merely defining a new method in Derived. Derived has no knowledge Base's implementation of func() since its declared as private. You won't get a compiler error when you define func() in Derived but that is because Derived does not know Base has an implementation of func(). To be clear: it would be incorrect to say you are overriding Base's implementation of func().
In addition to the already correct answer, consider this:
public class Private {
static class A {
public void doStuff() {
System.out.println(getStuff());
}
private String getStuff() {
return "A";
}
}
static class B extends A {
public String getStuff() {
return "B";
}
}
public static void main(String[] args) {
A a = new A();
a.doStuff();
a = new B();
a.doStuff();
B b = new B();
b.doStuff();
}
}
This will print
A
A
A
although B "overrides" getStuff(). As implementation of doStuff() is fixed to calling A#getStuff(), no polymorphism will be triggered.
Nope because if you do something like Base b = new Derived(); you still won't be able to call b.func(). What you're doing is called "hiding".
Since the method is private it is not visible to the other classes.Hence the derived class does not inherit this method.
So this is not the case of overriding
Method hiding will be happening here instead of overriding. like what happens in case of static.
Actually,you are not overriding.Before Java5
an overridden method's return type must match with parent class's method.
But Java 5 introduced a new facility called covariant return type.You can override a method with the same signature but returns a subclass of the object returned. In another words, a method in a subclass can return an object whose type is a subclass of the type returned by the method with the same signature in the superclass.
you can follow this thread :Can overridden methods differ in return type?
The private member of the base class cannot be access by anyone outside of the class and cannot be overridden. The function in the derive class is an independent function that can be access by anywhere.
The code would run the function in the derived class
A private method can never be over ridden. It is always hidden.
In your example - Derived class has one parent class private method and has its own function func. Both are different, and the func is not over ridden. Its a separate independent function.
If you create a new function in parent class calling parent class function, the parent func will be called, if parent class reference is used as opposed in the case of method over ridding
Note : An object defines the members which it has, and a reference defines which it can access
// Method Over ridding case
class Base{
public void func(){
System.out.println("Parent class");
};
public void func1(){
func();
}
}
class Derived extends Base{
public void func(){
System.out.println("Derived class");
}
}
class InheritDemo{
public static void main(String [] args){
Derived d = new Derived();
d.func1(); // Prints Derived class
Base b = new Derived();
b.func1(); // Prints Derived class - no matter parent reference is calling,as there as method is overridden - Check func1() is in parent class, but id doesn't call parent class func() as the compiler finds a func() method over ridden in derived class
}
}
// Method Hidding case - Private and static methods case
class Base{
private void func(){
System.out.println("Parent class");
};
public void func1(){
func()
}
}
class Derived extends Base{
public void func(){ // Is this a Method Overriding..????
System.out.println("Derived class");
}
}
class InheritDemo{
public static void main(String [] args){
Derived d = new Derived();
d.func1(); // Prints Derived class
Base b = new Derived();
b.func1();
// Prints Parent class - the reason is we are using the parent class reference, so compiler is looking for func() and it founds that there is one private class method which is available and is not over ridden, so it will call it. Caution - this won't happen if called using derived class reference.
b.func();
// this prints the Derived class - the compiler is looking func(), as Derived class has only one func() that it is implementing, so it will call that function.
}
}
Read comments in the below code snippet to find the answer.
Sources:
Definition reference:
Credits for the source code example(reference) from the book - "OCA Oracle Certified Associate Java SE 8 Programmer Study Guide Exam 1Z0-808 Book" from 'Jeanne Boyarsky' and 'Scott Selikoff'.
public class Deer {
public Deer() { System.out.print("Deer"); }
public Deer(int age) { System.out.print("DeerAge"); }
private boolean hasHorns() { return false; }
public static void main(String[] args) {
Deer deer = new Reindeer(5);
System.out.println(","+deer.hasHorns());// false is printed.
}
}
class Reindeer extends Deer {
public Reindeer(int age) { System.out.print("Reindeer"); }
private boolean hasHorns() { return true; } // Overriding possible, but is of no use in the below context.
// Below code is added by me for illustration purpose
public static void main(String[] args) {
Deer deer = new Reindeer(5);
//Below line gives compilation error.
//System.out.println(","+deer.hasHorns());
}
}