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

User · Answer

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.

User · Answer

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  gt  gt  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  gt  gt  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

User · Answer

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

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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

User · Answer

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   gt  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   gt  Inside start method of child

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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

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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

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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  gt  gt  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

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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

[java] Static Vs. Dynamic Binding in Java

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