Function Overloading

In this practical session we were introduced to function overloading in OOP. Using the concept of overloading we can design a group of functions with one name but different argument lists.   

#1. Problem statement:

We have to calculate the average of three numbers using the idea of function overloading.

Solution:

#include<iostream.h> #include<conio.h> int avg(int a,int b,int c); double avg(double a,double b,double c); double avg(int a,double b,double c); double avg(double a,double b,int c); void main(){       float x,y,z;       clrscr();       avg(1,2,3);       avg(6.5,7.5,5.5);       avg(3,6.5,5.0);       avg(6.5,5.0,3);       getch(); } int avg(int a,int b,int c){       int i;       i=(a+b+c)/3;       cout<<"the avg of 1,2,3 is="<<i<<endl;       return i; } double avg(double a,double b,double c){       float j;       j=(a+b+c)/3;       cout<<"the avg of 6.5,7.5,5.5 is="<<j<<endl;       return j; } double avg(int a,double b,double c){       float k;       k=(a+b+c)/3;       cout<<"the avg of 3,6.5,5.0 is="<<k<<endl;       return k; } double avg(double a,double b,int c){       float l;       l=(a+b+c)/3;       cout<<"the avg of 6.5,5.0,3 is="<<l<<endl;       return l; }

Remark:

I solved the program successfully,

#2.Problem statement:

We have to add two numbers and devide the result by the third one.

Solution:

#include<iostream.h> #include<conio.h> int avg(int a,int b,int c); double avg(double a,double b,double c); double avg(int a,double b,double c); double avg(double a,double b,int c); void main(){       float x,y,z;       clrscr();       avg(1,2,3);       avg(6.5,7.5,5.5);       avg(3,6.5,5.0);       avg(6.5,5.0,3);       getch(); } int avg(int a,int b,int c){       int i;       i=(a+b)/c;       cout<<"the result of adding 1 & 2 then dividing by 3 is="<<i<<endl;       return i; } double avg(double a,double b,double c){       float j;       j=(a+b)/c;       cout<<"the result of adding 6.5 & 7.5 then dividing by 5.5 is="<<j<<endl;       return j; } double avg(int a,double b,double c){       float k;       k=(a+b)/c;       cout<<"the result of adding 3 & 6.5 then dividing by 5 is="<<k<<endl;       return k; } double avg(double a,double b,int c){       float l;       l=(a+b)/c;       cout<<"the result of adding 6.5 & 5.0 then dividing by 3 is="<<l<<endl;       return l; }

Remark:

I solved the program successfully,

I solved the problems passing the actual arguments