Overloading Functions

Overloading Functions

Definition : Function overloading is a concept in which one can use many functions having same function name but can pass different number of parameters or different types of parameters.

Rules for Function Overloading

Following are the rules for function overloading ‒

1. The overloaded functions may differ by number of parameters.

2. The overloaded functions may differ by data types.

3. The same function name is used for various instances of function call.

 

For example, if there is a function sum which performs addition operation then one can use overloading functions like this ‒

int sum(int a,int b);

int sum(int a,int b,int c);

int sum(int a,int b,int c,int d);

Similarly the function overloading can be achieved like this ‒

int sum(int a,int b);

float sum(float a,float b);

double sum(double a,double b);

char sum(char a,char b);

That means we can handle different number of parameters or different types of parameters using the same function name.

Following program takes different number of parameters and performs the task of addition. Note that name of these functions is same and that is sum.

C++ Program

#include <iostream>

using namespace std;

class test {

  private:

        int a,b,c,d;

  public:

        int sum(int,int);

        int sum(int,int,int);

        int sum(int,int,int,int);

};

int test::sum(int a,int b)

{

        return (a+b);

}

int test::sum(int a,int b,int c)

{

        return (a+b+c);

}

int test::sum(int a,int b,int c,int d)

{

        return (a+b+c+d);

}

int main()

{

test obj;

int a, b, c, d,choice;

int result=0;

cout<<"\n\t\t Main Menu";

cout<<"\n\t 1.Addition of two numbers";

cout<<"\n\t 2.Addition of three numbers";

cout<<"\n\t 3.Addition of four numbers" <endl;

cout<<"\n Enter Your choice: ";

cin >> choice;

switch(choice)

{

        case 1: cout <<"\nEnter 2 numbers: ";

                cin >> a >> b;

                result = obj.sum(a,b);  // Overloading Functions

                break;

        case 2: cout <<"\nEnter 3 numbers: ";

                cin >> a >> b >> c;

                result = obj.sum(a,b,c); // Overloading Functions

                break;

        case 3: cout<<"\nEnter 4 numbers: ";

                cin >> a >>b>>c>> d;

                result = obj.sum(a,b,c,d); // Overloading Functions

                break;

        default: cout<<"\nWrong Choice";

                break;

}

cout<<"\n\nresult: << result << endl;

return 0;

}

Output

     Main Menu

1.Addition of two numbers

2.Addition of three numbers

3.Addition of four numbers

Enter Your choice: 2

Enter 3 numbers: 10 20 30

result: 60

• Now following program will take the parameters of different data types. This makes the function to overload.

C++ Program

#include <iostream>

using namespace std;

class test {

   public:

   int sum(int,int);

   float sum(float,float);

   double sum(double,double);

};

int test::sum(int a,int b)

{

   return (a+b);

}

float test::sum(float a,float b)

{

   return (a+b);

}

double test::sum(double a,double b)

{

   return (a+b);

}

int main()

{

test obj;

int choice;

int a,b;

float x,y;

double m,n;

double result=0;

cout<<"\n\t\t Main Menu";

cout<<"\n\t 1.Addition of two integer numbers";

cout<<"\n\t 2.Addition of float numbers ";

cout<<"\n\t 3.Addition of double numbers"<<endl;

cout<<"\n Enter Your choice: ";

cin >> choice;

switch(choice)

{

   case 1: cout <<"\nEnter 2 numbers: ";

      cin >> a >> b;

      result = obj.sum(a,b);

      break;

   case 2: cout <<"\nEnter 2 numbers: ";

      cin >> x >> y;

      result = obj.sum(x,y);

      break;

   case 3: cout <<"\nEnter 2 numbers:

      cin >> m >> n;

      result = obj.sum(m,n);

      break;

   default: cout <<"\nWrong Choice";

      break;

}

cout<<"\n\nresult: " << result << endl;

retum 0;

}

Output

Main Menu

1.Addition of two integer numbers

2.Addition of float numbers

3.Addition of double numbers

Enter Your choice: 2

Enter 2 numbers: 1.13   5.6

result: 6.73

Example: 1

 Write a C++ program that can take either two integers or two floating point. numbers and outputs the smallest number using class, function overloading.

Solution:

#include <iostream>

using namespace std;

class Test

{

public:

     int smallest(int,int);

     float smallest(float,float);

};

 int Test::smallest(int a,int b)

{

     if(a<b)

          return a;

     else

          return b;

}

float Test::smallest (float x,float y)

{

     if(x<y)

          return x;

     else

          return y;

}

int main()

{

Test obj;

class choice;

int a,b;

float x,y;

cout<<"\n Menu"<<endl;

cout<<"\n 1. Integers \n 2.Float "<<endl;

cout<<"\n Enter your choice: ";

cin>>choice;

switch(choice)

{

case 1:

     cout<<"\n Enter two numbers: "<<endl;

     cin>>a;

     cin>>b;

     cout<<"Smallest Number is :"<<obj.smallest(a,b);

     break;

case 2:

      cout<<"\n Enter two numbers: "<<endl;

     cin>>x;

     cin>>y;

     cout<<"Smallest Number is: "<<obj.smallest(x,y);

     break;

}

return 0;

}

Output

Menu

1. Integers

2. Float

Enter your choice: 1

Enter two numbers:

11

 7

Smallest Number is: 7

Example:2

 Design calculator using function overloading.

Solution:

#include<iostream>

#include<cstdlib>

using namespace std;

class Calculator

{

public:

int addition(int&,int);

float addition(float&, float);

int subtraction(int&,int);

float subtraction(float&, float);

int Multiplication(int&,int);

float Multiplication(float&, float);

int division(int&,int);

float division(float&, float);

};

void getdata(float& result, float temp);

float result,temp;

char sign;

int Calculator::addition(int& result,int temp)

{

     result=result+temp;

     return result;

}

float Calculator::addition(float& result, float temp)

{

      result=result+temp;

     return result;

}

int Calculator::subtraction(int& result,int temp)

{

     result=result‒temp;

     return result;

}

float Calculator::subtraction (float& result, float temp)

{

     result=result‒temp;

     return result;

}

int Calculator::Multiplication(int& result,int temp)

{

     result=result*temp;

     return result;

}

float Calculator::Multiplication(float& result, float temp)

{

     result=result*temp;

     return result;

}

int Calculator::division(int& result,int temp)

{

     result=result/temp;

     return result;

}

float Calculator::division (float& result, float temp)

{

     result=result/temp;

     return result;

}

void getdata(float& result, float temp)

{

     Calculator obj;

     char oper="\0';

     cout<<"\n\t\t Calculator Program";

     cout<<"\n Enter some expression";

     cout<<"\n First of all enter some number then enter an operator and then some number.";

     cout<<"\n For obtaining the result type = and then press Enter";

     cout<<"\n For Quitting the program type q or Q";

     cin >> result;

while(1)

{

cin >>oper;

if((oper=='q')||(oper=='Q'))

break;

if (oper= = '=')

{

     result;

      cout << result << endl;

}

else

 {

     cin >>temp://reading second operand for continuing the expression

}

switch(oper)

{

case '+':

     result=obj.addition(result,temp);

     result=result;

     break;

case ':

     result=obj.subtraction(result,temp);

     result=result;

     chat break;

case '*':

     result=obj.Multiplication(result,temp);

     result=result;

     break;

case '/':

     result=obj.division(result,temp);

     result=result;

     break;

}

}

}

void main()

{

     float val, first_num;

     clrscr();

     getdata(val, first_num);//function for entering expression

}

Output

Calculator Program

Enter some expression

First of all enter some number then enter an operator and then some number.

For obtaining the result type = and then press Enter

For Quitting the program type q or Q

2+3=

5

*7=

35

 q

Example: 3

Illustrate the concept of function overloading to find the maximum of two numbers.

Solution :

class test

{

public:

     int max(int,int);

     double max(double,double);

};

int test::max(int a,int b)

{

if(a>b)

      return a;

else

     return b;

}

double test::max(double a,double b)

{

if(a>b)

      return a;

else

     return b;

}

void main()

{

        test obj;

        cout<<"\n The maximum number is: "<<obj.max(10,20);

        cout<<"\n The maximum number is: "<<obj.max(11.11,2.22);

}

Example: 4

Write a C++ program to find area of the square, rectangle, circle using function overloading.

Solution :

#include<iostream>

using namespace std;

void area(int r)

{

     double a = 3.14*r*r;

     cout << "\n The area of circle is: " << a;

}

void area(int 1, int b)//overloaded function

{

     int a;

     a = 1*b;

     cout << "\n The area of rectangle is: " << a;

}

void area(double val)//overloaded function

{

     double a = val*val;

     cout << "\n The area of square is: << a;

}

void main()

{

int choice;

int radius, length, breadth;

double side;

char ans = 'y';

cout << "\n Program for Demonstrating function overloading";

do

{

     cout << "\n 1. Area of Square \n 2. Area of Rectangle \n 3. Area of Circle";

     cout << "\n Enter your choice: ";

     cin >> choice;

     switch (choice)

     {

          case 1:cout << "\n Enter side: ":

               cin >> side;

               area(side);

               break;

          case 2:cout << "\n Enter length: ";

               cin >> length;

               cout << "\n Enter breadth: ";

               cin >> breadth;

               area(length, breadth);

               break;

          case 3:cout << "\n Enter radius:

               cin >> radius;

               area(radius);

                break;

}

cout << "\n Do you want to continue?(y/n) ";

cin >> ans;

} while (ans= = 'y');

}

Output

Program for Demonstrating function overloading

1. Area of Square

2. Area of Rectangle

3. Area of Circle

Enter your choice: 1

Enter side: 11.11

The area of square is: 123.432

Do you want to continue?(y/n) y

 1. Area of Square

2. Area of Rectangle

3. Area of Circle

Enter your choice: 2

Enter length: 10

Enter breadth: 20

The area of rectangle is: 200

Do you want to continue?(y/n) y

1. Area of Square

2. Area of Rectangle

3. Area of Circle

Enter your choice: 3

Enter radius: 5

The area of circle is: 78.5

Do you want to continue? (y/n) n