Question

Please have a look at the following code

#include <iostream>
#include <iomanip>
#include <vector>
#include <string>

using namespace std;

void isPerfect(int);

vector<int> list;
string numList = "";


int main()
{
    cout << "Number" << setw(10) << "Divisors" << setw(15) << "Calculation";

    for(int i=1;i<=1000;i++)
    {
        isPerfect(i);
    }
}


void isPerfect(int number)
{

    int calc = 0;

    for(int i=1;i<=(number/2);i++)
    {
        if(number%i == 0)
        {
            list.push_back(i);
            //numList = numList +  string. + ",";
        }
    }

    for(size_t i=0;i<list.size();i++)
    {
        calc = list[i] + calc;
    }

    if(calc == number)
    {
        cout << number << setw(10) << numList << setw(15) << calc << endl;
    }
}

In here, I am trying to find the “Perfect Numbers” (perfect numbers – if the sum of it’s divisors, including 1, but not it self, is equal to the number, it is a perfect number. ex 6 )

But in here, everything is correct, but I am not getting any result, rather than printing what I have printed in main method.

Why is this happening? This is not a home work anyway. Please help.

Answer 1

Your list is in a global variable, so it also contains divisors of all previous numbers. Declare it locally in isPerfect.

As you’ll subsequently discover, the same goes for numList.