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Does static binding really happen for overloaded methods in Java? [duplicate]

I'm currently doing an assignment for one of my classes, and in it, I have to give examples, using Java syntax, of static and dynamic binding.
I understand the basic concept, that static binding happens at compile time and dynamic binding happens at runtime, but I can't figure out how they actually work specifically.
I found an example of static binding online that gives this example:
public static void callEat(Animal animal) {
System.out.println("Animal is eating");
}
public static void callEat(Dog dog) {
System.out.println("Dog is eating");
}
public static void main(String args[])
{
Animal a = new Dog();
callEat(a);
}
And that this would print "animal is eating" because the call to callEat uses static binding, but I'm unsure as to why this is considered static binding.
So far none of the sources I've seen have managed to explain this in a way that I can follow.

From Javarevisited blog post:
Here are a few important differences between static and dynamic binding:
Static binding in Java occurs during compile time while dynamic binding occurs during runtime.
private, final and static methods and variables use static binding and are bonded by compiler while virtual methods are bonded during runtime based upon runtime object.
Static binding uses Type (class in Java) information for binding while dynamic binding uses object to resolve binding.
Overloaded methods are bonded using static binding while overridden methods are bonded using dynamic binding at runtime.
Here is an example which will help you to understand both static and dynamic binding in Java.
Static Binding Example in Java
public class StaticBindingTest {
public static void main(String args[]) {
Collection c = new HashSet();
StaticBindingTest et = new StaticBindingTest();
et.sort(c);
}
//overloaded method takes Collection argument
public Collection sort(Collection c) {
System.out.println("Inside Collection sort method");
return c;
}
//another overloaded method which takes HashSet argument which is sub class
public Collection sort(HashSet hs) {
System.out.println("Inside HashSet sort method");
return hs;
}
}
Output: Inside Collection sort method
Example of Dynamic Binding in Java
public class DynamicBindingTest {
public static void main(String args[]) {
Vehicle vehicle = new Car(); //here Type is vehicle but object will be Car
vehicle.start(); //Car's start called because start() is overridden method
}
}
class Vehicle {
public void start() {
System.out.println("Inside start method of Vehicle");
}
}
class Car extends Vehicle {
#Override
public void start() {
System.out.println("Inside start method of Car");
}
}
Output: Inside start method of Car

Connecting a method call to the method body is known as Binding. As Maulik said "Static binding uses Type(Class in Java) information for binding while Dynamic binding uses Object to resolve binding." So this code :
public class Animal {
void eat() {
System.out.println("animal is eating...");
}
}
class Dog extends Animal {
public static void main(String args[]) {
Animal a = new Dog();
a.eat(); // prints >> dog is eating...
}
#Override
void eat() {
System.out.println("dog is eating...");
}
}
Will produce the result: dog is eating... because it is using the object reference to find which method to use. If we change the above code to this:
class Animal {
static void eat() {
System.out.println("animal is eating...");
}
}
class Dog extends Animal {
public static void main(String args[]) {
Animal a = new Dog();
a.eat(); // prints >> animal is eating...
}
static void eat() {
System.out.println("dog is eating...");
}
}
It will produce : animal is eating... because it is a static method, so it is using Type (in this case Animal) to resolve which static method to call. Beside static methods private and final methods use the same approach.

Well in order to understand how static and dynamic binding actually works? or how they are identified by compiler and JVM?
Let's take below example where Mammal is a parent class which has a method speak() and Human class extends Mammal, overrides the speak() method and then again overloads it with speak(String language).
public class OverridingInternalExample {
private static class Mammal {
public void speak() { System.out.println("ohlllalalalalalaoaoaoa"); }
}
private static class Human extends Mammal {
#Override
public void speak() { System.out.println("Hello"); }
// Valid overload of speak
public void speak(String language) {
if (language.equals("Hindi")) System.out.println("Namaste");
else System.out.println("Hello");
}
#Override
public String toString() { return "Human Class"; }
}
// Code below contains the output and bytecode of the method calls
public static void main(String[] args) {
Mammal anyMammal = new Mammal();
anyMammal.speak(); // Output - ohlllalalalalalaoaoaoa
// 10: invokevirtual #4 // Method org/programming/mitra/exercises/OverridingInternalExample$Mammal.speak:()V
Mammal humanMammal = new Human();
humanMammal.speak(); // Output - Hello
// 23: invokevirtual #4 // Method org/programming/mitra/exercises/OverridingInternalExample$Mammal.speak:()V
Human human = new Human();
human.speak(); // Output - Hello
// 36: invokevirtual #7 // Method org/programming/mitra/exercises/OverridingInternalExample$Human.speak:()V
human.speak("Hindi"); // Output - Namaste
// 42: invokevirtual #9 // Method org/programming/mitra/exercises/OverridingInternalExample$Human.speak:(Ljava/lang/String;)V
}
}
When we compile the above code and try to look at the bytecode using javap -verbose OverridingInternalExample, we can see that compiler generates a constant table where it assigns integer codes to every method call and byte code for the program which I have extracted and included in the program itself (see the comments below every method call)
By looking at above code we can see that the bytecodes of humanMammal.speak(), human.speak() and human.speak("Hindi") are totally different (invokevirtual #4, invokevirtual #7, invokevirtual #9) because the compiler is able to differentiate between them based on the argument list and class reference. Because all of this get resolved at compile time statically that is why Method Overloading is known as Static Polymorphism or Static Binding.
But bytecode for anyMammal.speak() and humanMammal.speak() is same (invokevirtual #4) because according to compiler both methods are called on Mammal reference.
So now the question comes if both method calls have same bytecode then how does JVM know which method to call?
Well, the answer is hidden in the bytecode itself and it is invokevirtual instruction set. JVM uses the invokevirtual instruction to invoke Java equivalent of the C++ virtual methods. In C++ if we want to override one method in another class we need to declare it as virtual, But in Java, all methods are virtual by default because we can override every method in the child class (except private, final and static methods).
In Java, every reference variable holds two hidden pointers
A pointer to a table which again holds methods of the object and a pointer to the Class object. e.g. [speak(), speak(String) Class object]
A pointer to the memory allocated on the heap for that object’s data e.g. values of instance variables.
So all object references indirectly hold a reference to a table which holds all the method references of that object. Java has borrowed this concept from C++ and this table is known as virtual table (vtable).
A vtable is an array like structure which holds virtual method names and their references on array indices. JVM creates only one vtable per class when it loads the class into memory.
So whenever JVM encounter with a invokevirtual instruction set, it checks the vtable of that class for the method reference and invokes the specific method which in our case is the method from a object not the reference.
Because all of this get resolved at runtime only and at runtime JVM gets to know which method to invoke, that is why Method Overriding is known as Dynamic Polymorphism or simply Polymorphism or Dynamic Binding.
You can read it more details on my article How Does JVM Handle Method Overloading and Overriding Internally.

The compiler only knows that the type of "a" is Animal; this happens at compile time, because of which it is called static binding (Method overloading). But if it is dynamic binding then it would call the Dog class method. Here is an example of dynamic binding.
public class DynamicBindingTest {
public static void main(String args[]) {
Animal a= new Dog(); //here Type is Animal but object will be Dog
a.eat(); //Dog's eat called because eat() is overridden method
}
}
class Animal {
public void eat() {
System.out.println("Inside eat method of Animal");
}
}
class Dog extends Animal {
#Override
public void eat() {
System.out.println("Inside eat method of Dog");
}
}
Output:
Inside eat method of Dog

There are three major differences between static and dynamic binding while designing the compilers and how variables and procedures are transferred to the runtime environment.
These differences are as follows:
Static Binding: In static binding three following problems are discussed:
Definition of a procedure
Declaration of a name(variable, etc.)
Scope of the declaration
Dynamic Binding: Three problems that come across in the dynamic binding are as following:
Activation of a procedure
Binding of a name
Lifetime of a binding

With the static method in the parent and child class: Static Binding
public class test1 {
public static void main(String args[]) {
parent pc = new child();
pc.start();
}
}
class parent {
static public void start() {
System.out.println("Inside start method of parent");
}
}
class child extends parent {
static public void start() {
System.out.println("Inside start method of child");
}
}
// Output => Inside start method of parent
Dynamic Binding :
public class test1 {
public static void main(String args[]) {
parent pc = new child();
pc.start();
}
}
class parent {
public void start() {
System.out.println("Inside start method of parent");
}
}
class child extends parent {
public void start() {
System.out.println("Inside start method of child");
}
}
// Output => Inside start method of child

All answers here are correct but i want to add something which is missing.
when you are overriding a static method, it looks like we are overriding it but actually it is not method overriding. Instead it is called method hiding. Static methods cannot be overridden in Java.
Look at below example:
class Animal {
static void eat() {
System.out.println("animal is eating...");
}
}
class Dog extends Animal {
public static void main(String args[]) {
Animal a = new Dog();
a.eat(); // prints >> animal is eating...
}
static void eat() {
System.out.println("dog is eating...");
}
}
In dynamic binding, method is called depending on the type of reference and not the type of object that the reference variable is holding
Here static bindinghappens because method hiding is not a dynamic polymorphism.
If you remove static keyword in front of eat() and make it a non static method then it will show you dynamic polymorphism and not method-hiding.
i found the below link to support my answer:
https://youtu.be/tNgZpn7AeP0

In the case of the static binding type of object determined at the compile-time whereas in
the dynamic binding type of the object is determined at the runtime.
class Dainamic{
void run2(){
System.out.println("dainamic_binding");
}
}
public class StaticDainamicBinding extends Dainamic {
void run(){
System.out.println("static_binding");
}
#Override
void run2() {
super.run2();
}
public static void main(String[] args) {
StaticDainamicBinding st_vs_dai = new StaticDainamicBinding();
st_vs_dai.run();
st_vs_dai.run2();
}
}

Because the compiler knows the binding at compile time. If you invoke a method on an interface, for example, then the compiler can't know and the binding is resolved at runtime because the actual object having a method invoked on it could possible be one of several. Therefore that is runtime or dynamic binding.
Your invocation is bound to the Animal class at compile time because you've specified the type. If you passed that variable into another method somewhere else, noone would know (apart from you because you wrote it) what actual class it would be. The only clue is the declared type of Animal.

Java inheritance: What is the difference between method overriding and method hiding? [duplicate]

This question already has answers here:
Overriding vs Hiding Java - Confused
(17 answers)
Closed last year.
According to the documentation
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.
While in case of static methods
If a subclass defines a static method with the same signature as a static method in the superclass, then the method in the subclass hides the one in the superclass.
So, I have tested the code example shown there with addition of more use cases:
The super class Animal:
public class Animal {
public static void testClassMethod() {
System.out.println("The static method in Animal");
}
public void testInstanceMethod() {
System.out.println("The instance method in Animal");
}
}
The subclass Cat:
public class Cat extends Animal {
public static void testClassMethod() {
System.out.println("The static method in Cat");
}
public void testInstanceMethod() {
System.out.println("The instance method in Cat");
}
public static void main(String[] args) {
Cat myCat = new Cat();
Animal myAnimal = myCat;
Animal animal = new Animal();
Animal.testClassMethod();
Cat.testClassMethod();
animal.testInstanceMethod();
myAnimal.testInstanceMethod();
}
}
The output here is:
The static method in Animal
The static method in Cat
The instance method in Animal
The instance method in Cat
So, I still see no actual difference between overriding the superclass instance method with subclass instance method with the same signature and overriding (hiding) the superclass static (class) method with subclass static method with the same signature.
What am I missing here?

Hiding it means that you can't call super.method() in the sub class's implementation.
So for example
class Cat extends Animal {
public static void testClassMethod() {
super.testClassMethod(); //this is not possible
System.out.println("The static method in Cat");
}
public void testInstanceMethod() {
super.testInstanceMethod(); //this is fine
System.out.println("The instance method in Cat");
}
}

You can call super() in an object method (not static), but you can not call super in a static method because there is no "super" object to call to.

How to call a method created in an inline class [duplicate]

This question already has answers here:
Calling newly defined method from anonymous class
(6 answers)
Closed 1 year ago.
I just came to know that there is a process called in-line class with the help of which I can create a new object and modify its class methods on the fly. I don't know if I can create new methods and variables inside the in-line class and use them. So I did this experiment. My IDE did not show any error while creating a new method inside the in-line class. Now I don't know how to access the newly created method. My doubt is, can I create a new method while creating an in-line class?(If yes then how?) Or, in-line class is only for overloading the existing methods?
public class Main {
public static void main(String[] args) {
Animals cat = new Animals("Cat") {
#Override
public void makeNoise() {
System.out.println("MEOW MEOW");
}
public void color() {
System.out.println("Color is: white");
}
};
cat.makeNoise();
cat.color(); //this is showing error
}
}
Animal class
class Animals {
private String name;
public Animals(String name) {
this.name = name;
}
public void makeNoise() {
System.out.println("Sound of the animal");
}
}

The easiest change: use var:
public static void main(String[] args) {
var cat = new Animals("Cat") {
#Override
public void makeNoise() {
System.out.println("MEOW MEOW");
}
public void color() {
System.out.println("Color is: white");
}
};
cat.makeNoise();
cat.color();
}
This now works, because cat isn't an Animals, it's inferred as a more specific type than that - it's a class that doesn't have an accessible name, so you can't write it as an explicit variable.
In fact, you were able to access anonymous class methods prior to the introduction of var, in a way:
new Animals("Cat") {
// ...
public void color() { ... }
}.color();
This worked prior to var, because the expression new Animals("Cat") { ... } has a more specific type that Animals. The problem is, you can only invoke those extra methods directly on the new instance creation expression, because you can only assign it to a variable of type Animals (or a superclass), thus preventing the compiler from accessing the specific class methods.
An alternative would be to declare it as a named class.
The most similar to what you have here would be a local class, although they are pretty rarely used (if known about at all), in my experience:
public static void main(String[] args) {
class Cat extends Animals {
Cat() { super("Cat"); }
// ...
public void color() { ... }
}
Cat cat = new Cat();
cat.color();
}
This is sort-of what the var approach does; it just gives the type an explicit name. This approach would be good if you wanted to create more than one instance of Cat in that method.
But there's not an obvious reason why this would need to be a local class: you could alternatively declare it as a nested or inner class, or, of course, a top-level class.

You cannot do that.. The first thing is Animals class has no method name Color.

Static binding in Java [duplicate]

I'm currently doing an assignment for one of my classes, and in it, I have to give examples, using Java syntax, of static and dynamic binding.
I understand the basic concept, that static binding happens at compile time and dynamic binding happens at runtime, but I can't figure out how they actually work specifically.
I found an example of static binding online that gives this example:
public static void callEat(Animal animal) {
System.out.println("Animal is eating");
}
public static void callEat(Dog dog) {
System.out.println("Dog is eating");
}
public static void main(String args[])
{
Animal a = new Dog();
callEat(a);
}
And that this would print "animal is eating" because the call to callEat uses static binding, but I'm unsure as to why this is considered static binding.
So far none of the sources I've seen have managed to explain this in a way that I can follow.

From Javarevisited blog post:
Here are a few important differences between static and dynamic binding:
Static binding in Java occurs during compile time while dynamic binding occurs during runtime.
private, final and static methods and variables use static binding and are bonded by compiler while virtual methods are bonded during runtime based upon runtime object.
Static binding uses Type (class in Java) information for binding while dynamic binding uses object to resolve binding.
Overloaded methods are bonded using static binding while overridden methods are bonded using dynamic binding at runtime.
Here is an example which will help you to understand both static and dynamic binding in Java.
Static Binding Example in Java
public class StaticBindingTest {
public static void main(String args[]) {
Collection c = new HashSet();
StaticBindingTest et = new StaticBindingTest();
et.sort(c);
}
//overloaded method takes Collection argument
public Collection sort(Collection c) {
System.out.println("Inside Collection sort method");
return c;
}
//another overloaded method which takes HashSet argument which is sub class
public Collection sort(HashSet hs) {
System.out.println("Inside HashSet sort method");
return hs;
}
}
Output: Inside Collection sort method
Example of Dynamic Binding in Java
public class DynamicBindingTest {
public static void main(String args[]) {
Vehicle vehicle = new Car(); //here Type is vehicle but object will be Car
vehicle.start(); //Car's start called because start() is overridden method
}
}
class Vehicle {
public void start() {
System.out.println("Inside start method of Vehicle");
}
}
class Car extends Vehicle {
#Override
public void start() {
System.out.println("Inside start method of Car");
}
}
Output: Inside start method of Car

Connecting a method call to the method body is known as Binding. As Maulik said "Static binding uses Type(Class in Java) information for binding while Dynamic binding uses Object to resolve binding." So this code :
public class Animal {
void eat() {
System.out.println("animal is eating...");
}
}
class Dog extends Animal {
public static void main(String args[]) {
Animal a = new Dog();
a.eat(); // prints >> dog is eating...
}
#Override
void eat() {
System.out.println("dog is eating...");
}
}
Will produce the result: dog is eating... because it is using the object reference to find which method to use. If we change the above code to this:
class Animal {
static void eat() {
System.out.println("animal is eating...");
}
}
class Dog extends Animal {
public static void main(String args[]) {
Animal a = new Dog();
a.eat(); // prints >> animal is eating...
}
static void eat() {
System.out.println("dog is eating...");
}
}
It will produce : animal is eating... because it is a static method, so it is using Type (in this case Animal) to resolve which static method to call. Beside static methods private and final methods use the same approach.

Well in order to understand how static and dynamic binding actually works? or how they are identified by compiler and JVM?
Let's take below example where Mammal is a parent class which has a method speak() and Human class extends Mammal, overrides the speak() method and then again overloads it with speak(String language).
public class OverridingInternalExample {
private static class Mammal {
public void speak() { System.out.println("ohlllalalalalalaoaoaoa"); }
}
private static class Human extends Mammal {
#Override
public void speak() { System.out.println("Hello"); }
// Valid overload of speak
public void speak(String language) {
if (language.equals("Hindi")) System.out.println("Namaste");
else System.out.println("Hello");
}
#Override
public String toString() { return "Human Class"; }
}
// Code below contains the output and bytecode of the method calls
public static void main(String[] args) {
Mammal anyMammal = new Mammal();
anyMammal.speak(); // Output - ohlllalalalalalaoaoaoa
// 10: invokevirtual #4 // Method org/programming/mitra/exercises/OverridingInternalExample$Mammal.speak:()V
Mammal humanMammal = new Human();
humanMammal.speak(); // Output - Hello
// 23: invokevirtual #4 // Method org/programming/mitra/exercises/OverridingInternalExample$Mammal.speak:()V
Human human = new Human();
human.speak(); // Output - Hello
// 36: invokevirtual #7 // Method org/programming/mitra/exercises/OverridingInternalExample$Human.speak:()V
human.speak("Hindi"); // Output - Namaste
// 42: invokevirtual #9 // Method org/programming/mitra/exercises/OverridingInternalExample$Human.speak:(Ljava/lang/String;)V
}
}
When we compile the above code and try to look at the bytecode using javap -verbose OverridingInternalExample, we can see that compiler generates a constant table where it assigns integer codes to every method call and byte code for the program which I have extracted and included in the program itself (see the comments below every method call)
By looking at above code we can see that the bytecodes of humanMammal.speak(), human.speak() and human.speak("Hindi") are totally different (invokevirtual #4, invokevirtual #7, invokevirtual #9) because the compiler is able to differentiate between them based on the argument list and class reference. Because all of this get resolved at compile time statically that is why Method Overloading is known as Static Polymorphism or Static Binding.
But bytecode for anyMammal.speak() and humanMammal.speak() is same (invokevirtual #4) because according to compiler both methods are called on Mammal reference.
So now the question comes if both method calls have same bytecode then how does JVM know which method to call?
Well, the answer is hidden in the bytecode itself and it is invokevirtual instruction set. JVM uses the invokevirtual instruction to invoke Java equivalent of the C++ virtual methods. In C++ if we want to override one method in another class we need to declare it as virtual, But in Java, all methods are virtual by default because we can override every method in the child class (except private, final and static methods).
In Java, every reference variable holds two hidden pointers
A pointer to a table which again holds methods of the object and a pointer to the Class object. e.g. [speak(), speak(String) Class object]
A pointer to the memory allocated on the heap for that object’s data e.g. values of instance variables.
So all object references indirectly hold a reference to a table which holds all the method references of that object. Java has borrowed this concept from C++ and this table is known as virtual table (vtable).
A vtable is an array like structure which holds virtual method names and their references on array indices. JVM creates only one vtable per class when it loads the class into memory.
So whenever JVM encounter with a invokevirtual instruction set, it checks the vtable of that class for the method reference and invokes the specific method which in our case is the method from a object not the reference.
Because all of this get resolved at runtime only and at runtime JVM gets to know which method to invoke, that is why Method Overriding is known as Dynamic Polymorphism or simply Polymorphism or Dynamic Binding.
You can read it more details on my article How Does JVM Handle Method Overloading and Overriding Internally.

The compiler only knows that the type of "a" is Animal; this happens at compile time, because of which it is called static binding (Method overloading). But if it is dynamic binding then it would call the Dog class method. Here is an example of dynamic binding.
public class DynamicBindingTest {
public static void main(String args[]) {
Animal a= new Dog(); //here Type is Animal but object will be Dog
a.eat(); //Dog's eat called because eat() is overridden method
}
}
class Animal {
public void eat() {
System.out.println("Inside eat method of Animal");
}
}
class Dog extends Animal {
#Override
public void eat() {
System.out.println("Inside eat method of Dog");
}
}
Output:
Inside eat method of Dog

There are three major differences between static and dynamic binding while designing the compilers and how variables and procedures are transferred to the runtime environment.
These differences are as follows:
Static Binding: In static binding three following problems are discussed:
Definition of a procedure
Declaration of a name(variable, etc.)
Scope of the declaration
Dynamic Binding: Three problems that come across in the dynamic binding are as following:
Activation of a procedure
Binding of a name
Lifetime of a binding

With the static method in the parent and child class: Static Binding
public class test1 {
public static void main(String args[]) {
parent pc = new child();
pc.start();
}
}
class parent {
static public void start() {
System.out.println("Inside start method of parent");
}
}
class child extends parent {
static public void start() {
System.out.println("Inside start method of child");
}
}
// Output => Inside start method of parent
Dynamic Binding :
public class test1 {
public static void main(String args[]) {
parent pc = new child();
pc.start();
}
}
class parent {
public void start() {
System.out.println("Inside start method of parent");
}
}
class child extends parent {
public void start() {
System.out.println("Inside start method of child");
}
}
// Output => Inside start method of child

All answers here are correct but i want to add something which is missing.
when you are overriding a static method, it looks like we are overriding it but actually it is not method overriding. Instead it is called method hiding. Static methods cannot be overridden in Java.
Look at below example:
class Animal {
static void eat() {
System.out.println("animal is eating...");
}
}
class Dog extends Animal {
public static void main(String args[]) {
Animal a = new Dog();
a.eat(); // prints >> animal is eating...
}
static void eat() {
System.out.println("dog is eating...");
}
}
In dynamic binding, method is called depending on the type of reference and not the type of object that the reference variable is holding
Here static bindinghappens because method hiding is not a dynamic polymorphism.
If you remove static keyword in front of eat() and make it a non static method then it will show you dynamic polymorphism and not method-hiding.
i found the below link to support my answer:
https://youtu.be/tNgZpn7AeP0

In the case of the static binding type of object determined at the compile-time whereas in
the dynamic binding type of the object is determined at the runtime.
class Dainamic{
void run2(){
System.out.println("dainamic_binding");
}
}
public class StaticDainamicBinding extends Dainamic {
void run(){
System.out.println("static_binding");
}
#Override
void run2() {
super.run2();
}
public static void main(String[] args) {
StaticDainamicBinding st_vs_dai = new StaticDainamicBinding();
st_vs_dai.run();
st_vs_dai.run2();
}
}

Because the compiler knows the binding at compile time. If you invoke a method on an interface, for example, then the compiler can't know and the binding is resolved at runtime because the actual object having a method invoked on it could possible be one of several. Therefore that is runtime or dynamic binding.
Your invocation is bound to the Animal class at compile time because you've specified the type. If you passed that variable into another method somewhere else, noone would know (apart from you because you wrote it) what actual class it would be. The only clue is the declared type of Animal.

method overriding using Interface

interface I
{
void show();
}
class A implements I
{
void show()
{
System.out.println("class A");
}
public static void main(String s[])
{
I i=new A();
i.show();
i.toString();
}
}
Q> As interface I does not contain the abstract method toString() but still The following code gets compiled. How?
when super class variable is used to refer sub class obj then compiler first searches the similar method in the super class if not found gives error.
here Interface does not contain the method toString().
ex=>
class A
{
void show()
{
System.out.println("show");
}
}
class B
{
void show()
{
System.out.println("show B");
}
void display()
{
System.out.println("display B");
}
public static void main(String s[])
{
A a=new B();
a.show(); //will execute
a.display(); //give error
}

All classes inherit from Object. Object has a toString.
To use any interface it must be backed by a actual class. So the Java compiler knows that it can use any method defined in java.lang.Object when dealing with an Interface.
To put it a slightly different way:
interface I { ... }
has an "magic"
interface I extends Object { ... }
So you can use Objects methods when detail with I. However you can not use any methods in the concrete class that do not appear in the interface. So to combine you two examples:
interface Car {
void drive();
}
class Convertible implements Car {
void drive() {}
void openRoof() {}
public static void main() {
Car porscheBoxster = new Convertible();
porscheBoxster.drive(); // OK - exists in interface
porscheBoxster.toString(); // OK - exists in java.lang.Object.
porscheBoxster.openRoof(); // Error. All we know is the porscheBoxster is of type Car.
// We don't know if it is a Convertible or not.
}
}

Every class in Java is an Object, thus, they are always able to run the following methods:
clone()
equals(Object)
finalize()
getClass()
hashCode()
notify()
notifyAll()
toString()
wait()
wait(long)
wait(long, int)

Because 'toString()' is in the class Object which every non-primitive data is derived from. So every object has this method.

In Java, every class you construct, inherits from the base class Object.
This means that your class by default will have a lot of useful methods, amongst others toString().