Static Vs. Dynamic Binding in Java

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

From Javarevisited blog post:

Here are a few important differences between static and dynamic binding:

1. Static binding in Java occurs during compile time while dynamic binding occurs during runtime.
2. 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.
3. Static binding uses Type (class in Java) information for binding while dynamic binding uses object to resolve binding.
4. 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

Answer 4

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

Answer 5

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

Answer 6

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