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Can dynamic polymorphism allow for greater flexibility than static polymorphism when writing in Java? [duplicate]

Can anyone provide a simple example that explains the difference between Dynamic and Static polymorphism in Java?

Polymorphism
1. Static binding/Compile-Time binding/Early binding/Method overloading.(in same class)
2. Dynamic binding/Run-Time binding/Late binding/Method overriding.(in different classes)
overloading example:
class Calculation {
void sum(int a,int b){System.out.println(a+b);}
void sum(int a,int b,int c){System.out.println(a+b+c);}
public static void main(String args[]) {
Calculation obj=new Calculation();
obj.sum(10,10,10); // 30
obj.sum(20,20); //40
}
}
overriding example:
class Animal {
public void move(){
System.out.println("Animals can move");
}
}
class Dog extends Animal {
public void move() {
System.out.println("Dogs can walk and run");
}
}
public class TestDog {
public static void main(String args[]) {
Animal a = new Animal(); // Animal reference and object
Animal b = new Dog(); // Animal reference but Dog object
a.move();//output: Animals can move
b.move();//output:Dogs can walk and run
}
}

Method overloading would be an example of static polymorphism
whereas overriding would be an example of dynamic polymorphism.
Because, in case of overloading, at compile time the compiler knows which method to link to the call. However, it is determined at runtime for dynamic polymorphism

Dynamic (run time) polymorphism is the polymorphism existed at run-time. Here, Java compiler does not understand which method is called at compilation time. Only JVM decides which method is called at run-time. Method overloading and method overriding using instance methods are the examples for dynamic polymorphism.
For example,
Consider an application that serializes and de-serializes different
types of documents.
We can have ‘Document’ as the base class and different document type
classes deriving from it. E.g. XMLDocument , WordDocument , etc.
Document class will define ‘ Serialize() ’ and ‘ De-serialize() ’
methods as virtual and each derived class will implement these
methods in its own way based on the actual contents of the documents.
When different types of documents need to be
serialized/de-serialized, the document objects will be referred by
the ‘ Document’ class reference (or pointer) and when the ‘
Serialize() ’ or ‘ De-serialize() ’ method are called on it,
appropriate versions of the virtual methods are called.
Static (compile time) polymorphism is the polymorphism exhibited at compile time. Here, Java compiler knows which method is called. Method overloading and method overriding using static methods; method overriding using private or final methods are examples for static polymorphism
For example,
An employee object may have two print() methods one taking no
arguments and one taking a prefix string to be displayed along with
the employee data.
Given these interfaces, when the print() method is called without any
arguments, the compiler, looking at the function arguments knows which function is meant to be called and it generates the object code
accordingly.
For more details please read "What is Polymorphism" (Google it).

Binding refers to the link between method call and method definition.
This picture clearly shows what is binding.
In this picture, “a1.methodOne()” call is binding to corresponding methodOne() definition and “a1.methodTwo()” call is binding to corresponding methodTwo() definition.
For every method call there should be proper method definition. This is a rule in java. If compiler does not see the proper method definition for every method call, it throws error.
Now, come to static binding and dynamic binding in java.
Static Binding In Java :
Static binding is a binding which happens during compilation. It is
also called early binding because binding happens before a program
actually runs
.
Static binding can be demonstrated like in the below picture.
In this picture, ‘a1’ is a reference variable of type Class A pointing to object of class A. ‘a2’ is also reference variable of type class A but pointing to object of Class B.
During compilation, while binding, compiler does not check the type of object to which a particular reference variable is pointing. It just checks the type of reference variable through which a method is called and checks whether there exist a method definition for it in that type.
For example, for “a1.method()” method call in the above picture, compiler checks whether there exist method definition for method() in Class A. Because ‘a1′ is Class A type. Similarly, for “a2.method()” method call, it checks whether there exist method definition for method() in Class A. Because ‘a2′ is also Class A type. It does not check to which object, ‘a1’ and ‘a2’ are pointing. This type of binding is called static binding.
Dynamic Binding In Java :
Dynamic binding is a binding which happens during run time. It is also
called late binding because binding happens when program actually is
running.
During run time actual objects are used for binding. For example, for “a1.method()” call in the above picture, method() of actual object to which ‘a1’ is pointing will be called. For “a2.method()” call, method() of actual object to which ‘a2’ is pointing will be called. This type of binding is called dynamic binding.
The dynamic binding of above example can be demonstrated like below.
Reference static-binding-and-dynamic-binding-in-java

method overloading is an example of compile time/static polymorphism because method binding between method call and method definition happens at compile time and it depends on the reference of the class (reference created at compile time and goes to stack).
method overriding is an example of run time/dynamic polymorphism because method binding between method call and method definition happens at run time and it depends on the object of the class (object created at runtime and goes to the heap).

In simple terms :
Static polymorphism : Same method name is overloaded with different type or number of parameters in same class (different signature). Targeted method call is resolved at compile time.
Dynamic polymorphism: Same method is overridden with same signature in different classes. Type of object on which method is being invoked is not known at compile time but will be decided at run time.
Generally overloading won't be considered as polymorphism.
From java tutorial page :
Subclasses of a class can define their own unique behaviors and yet share some of the same functionality of the parent class

Method Overloading is known as Static Polymorphism and also Known as Compile Time Polymorphism or Static Binding because overloaded method calls get resolved at compile time by the compiler on the basis of the argument list and the reference on which we are calling the method.
And Method Overriding is known as Dynamic Polymorphism or simple Polymorphism or Runtime Method Dispatch or Dynamic Binding because overridden method call get resolved at runtime.
In order to understand why this is so let's take an example of Mammal and Human class
class Mammal {
public void speak() { System.out.println("ohlllalalalalalaoaoaoa"); }
}
class Human extends Mammal {
#Override
public void speak() { System.out.println("Hello"); }
public void speak(String language) {
if (language.equals("Hindi")) System.out.println("Namaste");
else System.out.println("Hello");
}
}
I have included output as well as bytecode of in below lines of code
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
And by looking at above code we can see that the bytecodes of humanMammal.speak() , human.speak() and human.speak("Hindi") are totally different because the compiler is able to differentiate between them based on the argument list and class reference. And this is why Method Overloading is known as Static Polymorphism.
But bytecode for anyMammal.speak() and humanMammal.speak() is same because according to compiler both methods are called on Mammal reference but the output for both method calls is different because at runtime JVM knows what object a reference is holding and JVM calls the method on the object and this is why Method Overriding is known as Dynamic Polymorphism.
So from above code and bytecode, it is clear that during compilation phase calling method is considered from the reference type. But at execution time method will be called from the object which the reference is holding.
If you want to know more about this you can read more on How Does JVM Handle Method Overloading and Overriding Internally.

Polymorphism:
Polymorphism is the ability of an object to take on many forms. The most common use of polymorphism in OOP occurs when a parent class reference is used to refer to a child class object.
Dynamic Binding/Runtime Polymorphism :
Run time Polymorphism also known as method overriding. In this Mechanism by which a call to an overridden function is resolved at a Run-Time.
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
Static Binding /compile-time polymorphism:
Which method is to be called is decided at compile-time only.
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 metho

Static Polymorphism: is where the decision to resolve which method to accomplish, is determined during the compile time. Method Overloading could be an example of this.
Dynamic Polymorphism: is where the decision to choose which method to execute, is set during the run-time. Method Overriding could be an example of this.

Polymorphism refers to the ability of an object to behave differently for the same trigger.
Static polymorphism (Compile-time Polymorphism)
Static Polymorphism decides which method to execute during compile
time.
Method Overloading is an example of static polymorphism, and it is
requred to happens static polymorphism.
Static Polymorphism achieved through static binding.
Static Polymorphism happens in the same class.
Object assignment is not required for static polymorphism.
Inheritance not involved for static polymorphism.
Dynamic Polymorphism (Runtime Polymorphism)
Dynamic Polymorphism decides which method to execute in runtime.
Method Overriding is an example of dynamic polymorphism, and it is
requred to happens dynamic polymorphism.
Dynamic Polymorphism achieved through dynamic binding.
Dynamic Polymorphism happens between different classes.
It is required where a subclass object is assigned to super class
object for dynamic polymorphism.
Inheritance involved for dynamic polymorphism.

Compile time polymorphism(Static Binding/Early Binding): In static polymorphism, if we call a method in our code then which definition of that method is to be called actually is resolved at compile time only.
(or)
At compile time, Java knows which method to invoke by checking the method signatures. So, this is called compile-time polymorphism or static binding.
Dynamic Polymorphism(Late Binding/ Runtime Polymorphism): At run time, Java waits until runtime to determine which object is actually being pointed to by the reference. Method resolution was taken at runtime, due to that we call as run time polymorphism.

Consider the code below:
public class X
{
public void methodA() // Base class method
{
System.out.println ("hello, I'm methodA of class X");
}
}
public class Y extends X
{
public void methodA() // Derived Class method
{
System.out.println ("hello, I'm methodA of class Y");
}
}
public class Z
{
public static void main (String args []) {
//this takes input from the user during runtime
System.out.println("Enter x or y");
Scanner scanner = new Scanner(System.in);
String value= scanner.nextLine();
X obj1 = null;
if(value.equals("x"))
obj1 = new X(); // Reference and object X
else if(value.equals("y"))
obj2 = new Y(); // X reference but Y object
else
System.out.println("Invalid param value");
obj1.methodA();
}
}
Now, looking at the code you can never tell which implementation of methodA() will be executed, Because it depends on what value the user gives during runtime. So, it is only decided during the runtime as to which method will be called. Hence, Runtime polymorphism.

Method overloading is a compile time polymorphism, let's take an example to understand the concept.
class Person //person.java file
{
public static void main ( String[] args )
{
Eat e = new Eat();
e.eat(noodle); //line 6
}
void eat (Noodles n) //Noodles is a object line 8
{
}
void eat ( Pizza p) //Pizza is a object
{
}
}
In this example, Person has a eat method which represents that he can either eat Pizza or Noodles. That the method eat is overloaded when we compile this Person.java the compiler resolves the method call " e.eat(noodles) [which is at line 6] with the method definition specified in line 8 that is it method which takes noodles as parameter and the entire process is done by Compiler so it is Compile time Polymorphism.
The process of replacement of the method call with method definition is called as binding, in this case, it is done by the compiler so it is called as early binding.

Following on from Naresh's answer, dynamic polymorphism is only 'dynamic' in Java because of the presence of the virtual machine and its ability to interpret the code at run time rather than the code running natively.
In C++ it must be resolved at compile time if it is being compiled to a native binary using gcc, obviously; however, the runtime jump and thunk in the virtual table is still referred to as a 'lookup' or 'dynamic'. If C inherits B, and you declare B* b = new C(); b->method1();, b will be resolved by the compiler to point to a B object inside C (for a simple class inherits a class situation, the B object inside C and C will start at the same memory address so nothing is required to be done; it will be pointing at the vptr that they both use). If C inherits B and A, the virtual function table of the A object inside C entry for method1 will have a thunk which will offset the pointer to the start of the encapsulating C object and then pass it to the real A::method1() in the text segment which C has overridden. For C* c = new C(); c->method1(), c will be pointing to the outer C object already and the pointer will be passed to C::method1() in the text segment. Refer to: http://www.programmersought.com/article/2572545946/
In java, for B b = new C(); b.method1();, the virtual machine is able to dynamically check the type of the object paired with b and can pass the correct pointer and call the correct method. The extra step of the virtual machine eliminates the need for virtual function tables or the type being resolved at compile time, even when it could be known at compile time. It's just a different way of doing it which makes sense when a virtual machine is involved and code is only compiled to bytecode.

Which method is looked first by Compiler , Static or instance method when ClassName.method() is used?

I want to get proper understanding why below compilation error?
As per my understanding If i use Test.xyz() then compiler look for only static method not for instance method then why below compilation fail?
class Test {
public static void main(String arg[]) {
Test.xyz(10); // compilation fail
}
public void xyz(int i) {
}
public static void xyz(Integer i) {
}
}
Every one please suggest why compilation fail rather than other suggestions and how to use , I know all basic thing Autoboxing etc .

Why there is a compilation error
Compilation proceeds through different steps. Extracted from the JLS, following are the rules that explain why you got this error.
I skip the first step which is not relevant in your case. Everything happens in the same class.
Second step: Determine method signature
There may be more than one such method, in which case the most specific one is chosen. The descriptor (signature plus return type) of the most specific method is one used at run time to perform the method dispatch
The first phase (§15.12.2.2) performs overload resolution without permitting boxing or unboxing conversion, or the use of variable arity method invocation. If no applicable method is found during this phase then processing continues to the second phase.
From above comments, the method you invoke with Test.xyz(10); is the one that takes an int parameter:
public void xyz(int i) {}
But now, there is a third step: Choosing the appropriate method
If the method invocation has, before the left parenthesis, a MethodName of the form TypeName . Identifier, or if the method invocation, before the left parenthesis, has the form TypeName . NonWildTypeArguments Identifier, then the compile-time declaration must be static, or a compile-time error occurs.
Again, from the above comment, you invoke the method in a static form,
Test.xyz(10);
but unfortunately, the method chosen from the second step, is not static.
It's why an IDE like Eclipse will suggest to "Change 'xyz()' to static".
But as explained in my first answer (deleted), you can either call public void xyz(int i) {} on an instance of class Test, or call the static method with an Integer parameter: Test.xyz(Integer.valueOf(10));.
Both will work.

You have no return type here:
public static xyz(Integer i) {
}
This should be void, if there is nothing to return:
public static void xyz(Integer i) {
}
And also, you need to make the first method static too:
public static void xyz(int i) {
}
So it can be called out of the static main method. It is not possible to call non-static methods out of static methods. More detailed explanation on this: calling non-static method in static method in Java

This is how I did the static class.
public class test {
public static void main(String arg[]) {
xyz(10);
}
public static void xyz(int i) {
}
}

Ok, So here comes the concept of autoboxing in Java.
U wrote:
Test.xyz(10); // Here 10 is a primitive int and not a java.lang.Integer object.
But since you are invoking the xyz method directly via class name, clearly means you want to access the public static xyz(Integer) method of the class Test.
But what happens in the compilation process is, first your javac compiler checks for method signature that is to be accessed, and after that, it checks for it's access(public, private, protected, default) and non-access(final, static, etc) modifiers.
The same thing happened with this code.
The first thing Java did was, it checked for the existence of the method with the signature xyz(int) and not xyz(Integer) coz you've passed in 10 and not new Integer(10) as a parameter.
It found two methods,
1. xyz(int)
2. xyz(Integer)
Had xyz(int) not existed, it would have applied the concept of autoboxing(ie, automatically convert 10 to new Integer(10)) and selected xyz(Integer) to be executed. But since xyz(int) exists, it does not autobox 10 to new Integer(10) and selects xyz(int) instead of xyz(Integer).
Now since your compiler has selected xyz(int) to be executed, it checks for it's non-access modifiers. Now since the method xyz(int) is non-static, you are expected to access it with an object of Test class as follows:
new Test().xyz(10);
And still, if you want to access the static xyz(Integer) method, you may have to use:
Test.xyz(new Integer(10)); // You have to maunally autobox it.
Hope this helps.

Which constructor is called first while passing null in the class having overloaded constructor?

Below is the java class having 3 overloaded constructors :
public class Test {
public Test(Object i){
System.out.println("Object invoked");
}
public Test(String i){
System.out.println("String invoked");
}
public Test(int k){
System.out.println("Integer invoked");
}
public static void main(String[] args) throws Exception {
Test t = new Test(null);
}
}
If null value is passed while creating the new instance of class, which constructor will be invoked ? What is the reason ?

Java always chooses the most specific method (or constructor) that would be applicable for the argument you pass. In this case, that's the String constructor -- String is a subclass of Object.
Think about what would happen if you had
new Test("some string")
Both the Object and String constructors would be applicable here. After all, the argument is both an Object and a String. However, it is clear that the String constructor will be invoked because it is more specific than the Object constructor and still is applicable given the argument.
null is no different; these two constructors are still applicable, and the String constructor is still chosen over the Object one for the same reason.
Now, if there were two equally "specific" constructors present (e.g. if you had an Integer constructor) there would be a compilation error when you call Test(null).
This is outlined in more detail in JLS §15.12.2:
The second step searches the type determined in the previous step for member methods. This step uses the name of the method and the types of the argument expressions to locate methods that are both accessible and applicable, that is, declarations that can be correctly invoked on the given arguments.
There may be more than one such method, in which case the most specific one is chosen. The descriptor (signature plus return type) of the most specific method is one used at run time to perform the method dispatch.
The explicit process of determining which method is the most specific is outlined in JLS §15.12.2.5.

The answer is: Test(String) is invoked.
Why?
When determining which of a set of overloaded methods will be invoked, the Java compiler will attempt to match the most concrete type. And it will first attempt to match a signature before employing autoboxing. (#arshajii provided a perfect reference into the Java Language Spec on this)
Here, Object is the most abstract class in the type system. String subclasses from Object and is therefore more specific/concrete.
The logic behind this is that if you are overloading a method with a more specific-typed parameter, you're likely wanting to do more with that object (when you subclass, you typically add methods). If method signature determination worked the other way (i.e. the more abstractly-typed signature winning; here, Test(Object)), then none of the more concrete signatures would ever get called.

Ambiguity in function overloading in java

As return type does not play any role in function overloading,and compiler only check the unique existence of only those part of code which is used at calling time..
Then why not this code contain error
class Temp{
Temp(){
System.out.println("Default Constructor");
}
void Temp(){
S.o.p("HEll");
}
public static void main(String a[]){
new Temp();
}
}
Output=Default constructor...//
I am shocked there is NO COMPILE TIME ERROR as copiler only check only those part of code which is used at calling time ,means compiler need to check only unique existence of Temp() and there is no unique existence of Temp().
Kindly elaborate

Because void Temp() is a method and you should invoke it on instance. It is allowed to declare method with name of the class but invocation is different. You can call only constructor with new operator.
You can read more about Constructor in JLS#8.8. Constructor Declarations
Constructors are never invoked by method invocation expressions (§15.12).
Major difference here you should note is constructor do not have return types and definition of overloading includes return types.

One is a constructor, the other is not.

Compiler distinguishes between method and constructor call using keyword new. Only constructor can be after new keyword.

Constructors don't require a return type and void Temp() will be treated as method not constructor.
When you call void Temp();, constructor with no arguments will be invoked.

Classes, methods, and fields reside in different namespaces. Compiler always can determine what construct is used, and easily differentiates new Temp() (class) from o.Temp() (method) and o.Temp (field).

Using Eclipse, you may avoid this usage:
See the line above "Apply" button

Overloading of Math.sqrt : overloading method seems to hide the original one

trying to overload the java.lang.Math.sqrt static method for int type :
import static java.lang.Math.sqrt;
class Test
{
private static double sqrt(int n)
{
return sqrt(1.0 * n);
}
public static void main(String[] args)
{
System.out.println(sqrt(1));
}
}
an odd error arises :
Test.java:7: sqrt(int) in Test cannot be applied to (double)
return sqrt(1.0 * n);
^
1 error
But when explicitly referencing the java.lang.Math.sqrt method all is going fine :
class Test
{
private static double sqrt(int n)
{
return Math.sqrt(1.0 * n);
}
public static void main(String[] args)
{
System.out.println(sqrt(1));
}
}
The compiler used is the standard javac, version 1.6.0_16.
So the questions are :
Why is the compiler not able to resolve the overloading in the first case ?
Where does this behavior is specified in the java language specifications ?
Thanks in advance.

You can only overload methods in the same class. I.e. if you import a static method of another class and then define your own method with the same name, there will be no overload resolution. The imported version of the method will simply be ignored.
Regarding where this behavior is specified: The language specification defines overloading like this:
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.
Note that it says "two methods of a class". So methods imported from another class are simply not considered.
So since your definition is not an overload of Math.sqrt, it shadows it as per section 6.3.1 of the definition:
A declaration d of a method named n shadows the declarations of any other methods named n that are in an enclosing scope at the point where d occurs throughout the scope of d.

That's quite normal.
By writing this code, you didn't overload the Math.sqrt method in any way, you simply defined a Test.sqrt(int) as an argument.
Then there is the question of your error message.
When defining the Test.sqrt method, you overload (in this class) the static import you made.
So, when calling sqrt(1.0 * n), the compiler consider you try to call Test.sqrt(int) with a double, swhich is obviously not possible.

You're not actually overloading. Things are being hung up over an ambiguity; the compiler sees return sqrt(1.0 * n);, it assumes you're talking about the sqrt(int) function in class Test, not the one in lang.Math. This is preferable behavior; after all, you probably don't want to have to specify class.method every time you're trying to call a member function. And the compiler doesn't resolve it is because in your particular situation, it assumes you've made a mistake and doesn't realize you intended to call the sqrt in lang.Math.

Yes, sqrt(int) in Test cannot be applied to (double)
Rename your function with different name, if you wish to call Math.sqrt:
private static double mysqrt(int n)
{
return sqrt(1.0 * n);
}

In the first call the method referred is the one which you have return which expect int to be its parameter but you are sending double to it. so it gives a compiler error.
But with the second call you are referring to Math.sqrt() which works fine.