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C++ program to check if the number is abundant or not using class
Submitted by Shubh Pachori, on October 04, 2022
Problem statement
Given a number, we have to check whether it is abundant number or not using class and object approach.
Example:
Input:
Enter Number : 30
Output:
It is an Abundant Number
C++ code to check if the number is abundant or not using the class and object approach
#include <iostream>
using namespace std;
// create a class
class Abundant {
// private data member
private:
int number;
// public member functions
public:
// getNumber() function to insert number
void getNumber() {
cout << "Enter Number : ";
cin >> number;
}
// isAbundant() function to check if the number
// is abundant is or not
void isAbundant() {
// initialising int type variables to perform operations
int index, sum = 0;
// for loop to check the number
for (index = 1; index < number; index++) {
// if condition to check if the number
// is a multiple of the index
if (number % index == 0) {
sum = sum + index;
}
}
if (sum > number) {
cout << "It is an Abundant Number";
} else {
cout << "It is not an Abundant Number";
}
}
};
int main() {
// create an object
Abundant A;
// calling getNumber() function to
// insert the number
A.getNumber();
// calling isAbundant() function to check the
// number if it is a abundant number
A.isAbundant();
return 0;
}
Output
Enter Number : 30
It is an Abundant Number
Explanation
In the above code, we have created a class Abundant, one int type data member number to store the number, and public member functions getNumber() and isAbundant() to store and to check the given integer number is a abundant number or not.
In the main() function, we are creating an object A of class Abundant, reading an integer number by the user using the getNumber() function, and finally calling the isAbundant() member function to check the given integer number. The isAbundant() function contains the logic to given check the number and printing the result.
C++ Class and Object Programs (Set 2) »