undefined | unit 3 polymorphism

OOP

UNIT 3

Polymorphism

What is polymorphism explain with examples?

The term "Polymorphism" is the combination of "poly" + "morphs" which means many forms. It is a greek word. In object-oriented programming, we use 3 main concepts: inheritance, encapsulation, and polymorphism.

Real Life Example Of Polymorphism

Let's consider a real-life example of polymorphism. A lady behaves like a teacher in a classroom, mother or daughter in a home and customer in a market. Here, a single person is behaving differently according to the situations.

There are two types of polymorphism in C++:

Compile time polymorphism: The overloaded functions are invoked by matching the type and number of arguments. This information is available at the compile time and, therefore, compiler selects the appropriate function at the compile time. It is achieved by function overloading and operator overloading which is also known as static binding or early binding. Now, let's consider the case where function name and prototype is same.

class A // base class declaration.

{

int a;

public:

void display()

{

cout<< "Class A ";

}

};

class B : public A // derived class declaration.

{

int b;

public:

void display()

{

cout<<"Class B";

}

};

In the above case, the prototype of display() function is the same in both the base and derived class. Therefore, the static binding cannot be applied. It would be great if the appropriate function is selected at the run time. This is known as run time polymorphism.

Run time polymorphism: Run time polymorphism is achieved when the object's method is invoked at the run time instead of compile time. It is achieved by method overriding which is also known as dynamic binding or late binding.

2. Differences b/w compiles time and run time polymorphism.

Compile time polymorphism	Run time polymorphism
The function to be invoked is known at the compile time.	The function to be invoked is known at the run time.
It is also known as overloading, early binding and static binding.	It is also known as overriding, Dynamic binding and late binding.
Overloading is a compile time polymorphism where more than one method is having the same name but with the different number of parameters or the type of the parameters.	Overriding is a run time polymorphism where more than one method is having the same name, number of parameters and the type of the parameters.
It is achieved by function overloading and operator overloading.	It is achieved by virtual functions and pointers.
It provides fast execution as it is known at the compile time.	It provides slow execution as it is known at the run time.
It is less flexible as mainly all the things execute at the compile time.	It is more flexible as all the things execute at the run time.

3. Give example of run time polymorphism

Let's see a simple example of run time polymorphism in C++.

// an example without the virtual keyword.

#include <iostream>

using namespace std;

class Animal {

public:

void eat(){

cout<<"Eating...";

}

};

class Dog: public Animal

{

public:

void eat()

{ cout<<"Eating bread...";

}

};

int main(void) {

Dog d = Dog();

d.eat();

return 0;

}

Output:

Eating bread...

4. Give example of run time polymorphism in C++ where we are having two derived classes.

Let's see another example of run time polymorphism in C++ where we are having two derived classes.

// an example with virtual keyword.

#include <iostream>

using namespace std;

class Shape { // base class

public:

virtual void draw(){ // virtual function

cout<<"drawing..."<<endl;

}

};

class Rectangle: public Shape // inheriting Shape class.

{

public:

void draw()

{

cout<<"drawing rectangle..."<<endl;

}

};

class Circle: public Shape // inheriting Shape class.

{

public:

void draw()

{

cout<<"drawing circle..."<<endl;

}

};

int main(void) {

Shape *s; // base class pointer.

Shape sh; // base class object.

Rectangle rec;

Circle cir;

s=&sh;

s->draw();

s=&rec;

s->draw();

s=?

s->draw();

}

Output:

drawing...

drawing rectangle...

drawing circle...

5. Explain early and late binding with examples

In this section we will see what is early binding and what is late binding in C++. The binding means the process of converting identifiers into addresses. For each variables and functions this binding is done. For functions it is matching the call with the right function definition by the compiler. The binding is done either at compile time or at runtime.

Early Binding

This is compile time polymorphism. Here it directly associates an address to the function call. For function overloading it is an example of early binding.

Example

#include<iostream>

using namespace std;

class Base {

public:

void display() {

cout<<" In Base class" <<endl;

}

};

class Derived: public Base {

public:

void display() {

cout<<"In Derived class" << endl;

}

};

int main(void) {

Base *base_pointer = new Derived;

base_pointer->display();

return 0;

}

Output

In Base class

Late Binding

This is run time polymorphism. In this type of binding the compiler adds code that identifies the object type at runtime then matches the call with the right function definition. This is achieved by using virtual function.

Example

#include<iostream>

using namespace std;

class Base {

public:

virtual void display() {

cout<<"In Base class" << endl;

}

};

class Derived: public Base {

public:

void display() {

cout<<"In Derived class" <<endl;

}

};

int main() {

Base *base_pointer = new Derived;

base_pointer->display();

return 0;

}

Output

In Derived class

6. What is virtual function and pure virtual function?

Virtual Function

A virtual function is a function in a base class that is declared using the keyword virtual. Defining in a base class a virtual function, with another version in a derived class, signals to the compiler that we don't want static linkage for this function.

What we do want is the selection of the function to be called at any given point in the program to be based on the kind of object for which it is called. This sort of operation is referred to as dynamic linkage, or late binding.

Pure Virtual Functions

It is possible that you want to include a virtual function in a base class so that it may be redefined in a derived class to suit the objects of that class, but that there is no meaningful definition you could give for the function in the base class.

We can change the virtual function area() in the base class to the following −

class Shape {

protected:

int width, height;

public:

Shape(int a = 0, int b = 0) {

width = a;

height = b;

}

// pure virtual function

virtual int area() = 0;

};

The = 0 tells the compiler that the function has no body and above virtual function will be called pure virtual function.

7. Explain function overloading in detail with some examples

C++ Function Overloading

Function Overloading is defined as the process of having two or more function with the same name, but different in parameters is known as function overloading in C++. In function overloading, the function is redefined by using either different types of arguments or a different number of arguments. It is only through these differences compiler can differentiate between the functions.

The advantage of Function overloading is that it increases the readability of the program because you don't need to use different names for the same action.

C++ Function Overloading Example

Let's see the simple example of function overloading where we are changing number of arguments of add() method.

// program of function overloading when number of arguments vary.

#include <iostream>

using namespace std;

class Cal {

public:

static int add(int a,int b){

return a + b;

}

static int add(int a, int b, int c)

{

return a + b + c;

}

};

int main(void) {

Cal C; // class object declaration.

cout<<C.add(10, 20)<<endl;

cout<<C.add(12, 20, 23);

return 0;

}

Output:

Let's see the simple example when the type of the arguments vary.

// Program of function overloading with different types of arguments.

#include<iostream>

using namespace std;

int mul(int,int);

float mul(float,int);

int mul(int a,int b)

{

return a*b;

}

float mul(double x, int y)

{

return x*y;

}

int main()

{

int r1 = mul(6,7);

float r2 = mul(0.2,3);

std::cout << "r1 is : " <<r1<< std::endl;

std::cout <<"r2 is : " <<r2<< std::endl;

return 0;

}

Output:

r1 is : 42

r2 is : 0.6

Function Overloading and Ambiguity

When the compiler is unable to decide which function is to be invoked among the overloaded function, this situation is known as function overloading.

When the compiler shows the ambiguity error, the compiler does not run the program.

Causes of Function Overloading:

Type Conversion.

Function with default arguments.

Function with pass by reference.

Type Conversion:

Let's see a simple example.

#include<iostream>

using namespace std;

void fun(int);

void fun(float);

void fun(int i)

{

std::cout << "Value of i is : " <<i<< std::endl;

}

void fun(float j)

{

std::cout << "Value of j is : " <<j<< std::endl;

}

int main()

{

fun(12);

fun(1.2);

return 0;

}

The above example shows an error "call of overloaded 'fun(double)' is ambiguous". The fun(10) will call the first function. The fun(1.2) calls the second function according to our prediction. But, this does not refer to any function as in C++, all the floating point constants are treated as double not as a float. If we replace float to double, the program works. Therefore, this is a type conversion from float to double.

Function with Default Arguments

Let's see a simple example.

#include<iostream>

using namespace std;

void fun(int);

void fun(int,int);

void fun(int i)

{

std::cout << "Value of i is : " <<i<< std::endl;

}

void fun(int a,int b=9)

{

std::cout << "Value of a is : " <<a<< std::endl;

std::cout << "Value of b is : " <<b<< std::endl;

}

int main()

{

fun(12);

return 0;

}

The above example shows an error "call of overloaded 'fun(int)' is ambiguous". The fun(int a, int b=9) can be called in two ways: first is by calling the function with one argument, i.e., fun(12) and another way is calling the function with two arguments, i.e., fun(4,5). The fun(int i) function is invoked with one argument. Therefore, the compiler could not be able to select among fun(int i) and fun(int a,int b=9).

Function with pass by reference

Let's see a simple example.

#include <iostream>

using namespace std;

void fun(int);

void fun(int &);

int main()

{

int a=10;

fun(a); // error, which f()?

return 0;

}

void fun(int x)

{

std::cout << "Value of x is : " <<x<< std::endl;

}

void fun(int &b)

{

std::cout << "Value of b is : " <<b<< std::endl;

}

The above example shows an error "call of overloaded 'fun(int&)' is ambiguous". The first function takes one integer argument and the second function takes a reference parameter as an argument. In this case, the compiler does not know which function is needed by the user as there is no syntactical difference between the fun(int) and fun(int &).

8. Explain Operators Overloading in detail with some examples

Operator overloading is a compile-time polymorphism in which the operator is overloaded to provide the special meaning to the user-defined data type. Operator overloading is used to overload or redefines most of the operators available in C++. It is used to perform the operation on the user-defined data type. For example, C++ provides the ability to add the variables of the user-defined data type that is applied to the built-in data types.

The advantage of Operators overloading is to perform different operations on the same operand.

Operator that cannot be overloaded are as follows:

Scope operator (::)

Sizeof

member selector(.)

member pointer selector(*)

ternary operator(?:)

Syntax of Operator Overloading

return_type class_name : : operator op(argument_list)

{

// body of the function.

}

Where the return type is the type of value returned by the function.

class_name is the name of the class.

operator op is an operator function where op is the operator being overloaded, and the operator is the keyword.

Rules for Operator Overloading

Existing operators can only be overloaded, but the new operators cannot be overloaded.

The overloaded operator contains atleast one operand of the user-defined data type.

We cannot use friend function to overload certain operators. However, the member function can be used to overload those operators.

When unary operators are overloaded through a member function take no explicit arguments, but, if they are overloaded by a friend function, takes one argument.

When binary operators are overloaded through a member function takes one explicit argument, and if they are overloaded through a friend function takes two explicit arguments.

C++ Operators Overloading Example

Let's see the simple example of operator overloading in C++. In this example, void operator ++ () operator function is defined (inside Test class).

// program to overload the unary operator ++.

#include <iostream>

using namespace std;

class Test

{

private:

int num;

public:

Test(): num(8){}

void operator ++() {

num = num+2;

}

void Print() {

cout<<"The Count is: "<<num;

}

};

int main()

{

Test tt;

++tt; // calling of a function "void operator ++()"

tt.Print();

return 0;

}

Output:

The Count is: 10

Let's see a simple example of overloading the binary operators.

// program to overload the binary operators.

#include <iostream>

using namespace std;

class A

{

int x;

public:

A(){}

A(int i)

{

x=i;

}

void operator+(A);

void display();

};

void A :: operator+(A a)

{

int m = x+a.x;

cout<<"The result of the addition of two objects is : "<<m;

}

int main()

{

A a1(5);

A a2(4);

a1+a2;

return 0;

}

Output:

The result of the addition of two objects is : 9

9. Explain Overloading Unary Operators with some examples

Member operator function takes no parameter for unary operator and is implicitly passed through “this” pointer.

Overloading Unary Minus(–)

 Program

Write a program in C++ to overload unary minus(-).

Solution :

#include<iostream>

using namespace std ;

class Example

{

int x,y,z;

public:

Example(int p,int q,int r)

{

x=p;

y=q;

z=r;

}

void display()

{

cout<<x<<” ”<<y<<” ”<<z;

}

void operator-()

{

x=-x;

y=-y;

z=-z;

}

};

int main()

{

Example obj (10,-20,30);

cout<<”Original numbers are”

obj.display();

- obj;

obj.display();

return 0;

}

//Defines class Example

//Parameterized constructor

// Step 1 Operator function to overload minus

// Step 2 Invokes operator function

cout<<”Numbers after applying unary minus”

Explanation

This program overloads unary – on data members of class Example.

It defines operator function to overload unary – so that it can be applied to object. See step 1.

void operator-()

{

x=-x;

y=-y;

z=-z;

}

//Operator function to overload minus

This operator function is invoked by applying unary – on object ‘obj’. See step 2.

- obj;

cout<<”Numbers after applying unary minus”

//Invokes operator function

Output

Original numbers are

10 -20 30

Numbers after applying unary minus

-10 20 -30

10. Write C++ program to demonstrate use of binary (“+”) operator overloading to add two complex numbers using member function.

#include<iostream>

using namespace std ;

class Demo

{

int real,imag;

public:

Demo(int p, int q)

{

real=p;

imag=q;

}

void display()

{

cout<<real<<” +j”<<imag<<endl;

}

Demo operator+(Demo X);

};

Demo Demo::operator+(Demo X)

{

Demo D;

D.real=real+X.real;

D.imag=imag+D.imag;

return D

}

int main()

{

Demo obj1(10,20);

Demo obj2(5,5);

Demo obj3(0,0);

cout<<”Addition is”

obj3= obj1+ obj2;

obj3.display();

return 0 ;

}

//……1overloads +

//.2 Defines operator function to overload +

//Defines operator function to overload -

//…..3

//Invokes overloaded binary +

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