C++ Perfect Number With Nested Loops Issue

Question

What I am trying to do is search for a perfect number. A perfect number is a number that is the sum of all its divisors, such as 6 = 1+2+3.

Basically what I do here is ask for 2 numbers and find a perfect number between those two numbers. I have a function that tests for divisibility and 2 nested loops.

My issue is that I don't get any result. I've revised it & can't seem to find anything wrong. The compiler doesn't shoot out any errors.

What can be wrong?

#include <iostream>
using namespace std;

bool isAFactor(int, int);

int main()
{

int startval;
int endval;
int outer_loop;
int inner_loop;
int perfect_number = 0;

cout << "Enter Starting Number: ";
cin >> startval;
cout << "Enter Ending Number: ";
cin >> endval;

for(outer_loop = startval; outer_loop <= endval; outer_loop++)
{
    for(inner_loop = 1; inner_loop <= outer_loop; inner_loop++)
    {
        if (isAFactor(outer_loop, inner_loop) == true)
        {
            inner_loop += perfect_number;
        }
    }

if (perfect_number == outer_loop)
{
    cout << perfect_number << " is a perfect number." << endl;
}

else
{
    cout << "There is no perfect number." << endl;
}

}

system("PAUSE");
return 0;
}

bool isAFactor(int outer, int inner)
{
if (outer % inner == 0)
{
    return true;
}

else
{
    return false;
}

You are reading a string into an int value. That doesn't work. You have to read into a character array, and convert the first character(s) into an integer. — Seth Battin, Oct 21 '12 at 01:37
@SethBattin: try the questioner's code, you'll see that part of it is fine. — Steve Jessop, Oct 21 '12 at 01:43

score 2 · Accepted Answer · answered Oct 21 '12 at 01:36

inner_loop += perfect_number; should be perfect_number += inner_loop;.

There are other issues -- you need to reset perfect_number to zero in each outer loop, and you should presumably print the message "There is no perfect number." if none of the numbers in range is perfect, rather than printing it once for every number in range that is not perfect.

I'd advise that you rename perfect_number to sum_of_factors, outer_loop to candidate_perfect_number and inner_loop to candidate_factor, or similar.

Also, `perfect_number` will need to get reset each time through the outer loop. — Tom W, Oct 21 '12 at 01:38

Rustam Kichinsky · Answer 2 · 2012-10-21T01:41:18.593

0

after the if statement:

cout << perfect_number;
cout << outer_loop;

if (perfect_number == outer_loop)
{
    cout << perfect_number << " is a perfect number." << endl;
}

and see what values they have

Updated:

What is the value of your endval? is 0?, and thats why the loop ends so early

edited Oct 21 '12 at 01:41

answered Oct 21 '12 at 01:35

Rustam Kichinsky

119
1
7

score 0 · Answer 3 · answered Oct 21 '12 at 01:43

0

Oh, so many issues.

The variable perfect_number never changes. Did your compiler flag this?
The outer loop will be one more than the ending value when it exits; did you know this?
You don't need to compare bool values to true or false.
You could simplify the isAFactor function to return (outer % inner) == 0;.
You could replace the call to isAFactor with the expression ((outer % inner) == 0).

answered Oct 21 '12 at 01:43

Thomas Matthews

56,849
17
98
154

(1) is an error, (2) is fine (at least, provided endval != INT_MAX, that's a nasty corner case), (3) and (4) are redundant code that should be cleaned up but don't stop the program working, (5) is a matter of whether, philosophically speaking, you feel that subroutines are a good thing :-) Actually I think this code could benefit from *more* different functions rather than fewer. – Steve Jessop Oct 21 '12 at 01:46
Btw, what compiler option do I need to get a warning for `perfect_number` not being modified? That's a pretty fussy compiler, to nag you to mark it `const`. Why warn for `perfect_number` in `main` and not for (say) `outer` in `isAFactor` being non-const but never modified? – Steve Jessop Oct 21 '12 at 01:51

C++ Perfect Number With Nested Loops Issue

3 Answers3