Prime Factor Fxn

I need some help with the prime_factor fxn. I think I'm just missing something easy that I'm not seeing right now. Anyways, it's supposed to output the prime factors of an input number.

When I input 12, it outputs the desired results; however, if I enter in another nonprime number, it doesn't output the desired results. For example, if I input 6, it only says that 2 is a factor, but does not say 3 is a factor.

Ex. Output:

Enter an integer < 1000 -> 12

The prime factorization of the number 12 is:
2 is a factor
2 is a factor
3 is a factor
______________________

Enter an integer < 1000 -> 6

The prime factorization of the number 6 is:
2 is a factor

// It doesn't output that 3 is also a factor for the number 6.

Here's an example main to go with the program, followed by the function I need help with:

Code:

#include <iostream>
#include <string>
#include <cmath>
using namespace std;

const string SPACE_STR = "   ";

void prime_factor(int number);
bool is_prime(int number);

int main() {
   int number;

   cout << "Enter a number < 1000 -> ";
   cin >> number;

   prime_factor(number);

   return 0;
}

// Function I need help with.

void prime_factor(int number)
{
   bool isPrime = is_prime(number);
   int prime = number;
   int i = 2, j;
   double squareRoot = sqrt(static_cast<double>(number));
	int count = 0;
   
   cout << "The prime factorization of the number " << number << " is:" << endl;
	
	if(isPrime)
      cout << SPACE_STR << number << " is a prime number." << endl;
   else {
      while((prime > 0) && (i <= squareRoot)) {
         if((prime % i) == 0) {
				count++;
				for(j = 0; j < count; j++)
					cout << SPACE_STR;
            cout << i << " is a factor" << endl;
            prime /= i;
			} else         
         	i++;
      }
   }
}

bool is_prime(int number)
{
   int i;
   
   for(i = 2; i < number; i++) {
      if((number % i) == 0)
         return false;
   }
   
   return true;
}

your C++ code seems alright, but you have a logic flaw.

if number is 6
then sqrt of 6 is less than 3 and greater than 2 so squareRoot is something between 2 and 3, although I don't know exactly what it is.

6 is not prime so you end up going into this loop

Code:

while((prime > 0) && (i <= squareRoot)) 
{
         if((prime % i) == 0) 
        {
            cout << i << " is a factor" << endl;
            prime /= i;
        } else         
            i++;
}

the initial time through the loop prime = 6, i = 2;
prime % 2 equals 0 so the output is done and
prime is divided by 2 to get 3

then you go back to the while conditionals a second time
prime is 3 so it is > 0 and i is still 2 so it is <= squareRoot. Alls ok so far and you go into the if conditional where
3 % 2 is not 0 so the if conditional is false and you end up at else where i is increased from 2 to 3.

then you go back to the while conditionals a third time.
prime is still 3 so it is > 0 but now i is 3 and it is greater than squareRoot so the while conditional fails and you don't get to the if(prime % i == 0) part that would say 3 is prime and therefore 3 will not be recognized as a prime factor and it won't be printed out for you.

There are several ways to correct this logic so I'll let you have at it.

wow, I just changed the i <= squareRoot to i <= number and it works. It makes more sense now too. I got confused at where nonprime numbers have a factor below their square root.

Thanks for explaining it for me.