Extra "0" in output when factorizing a number - c++

Question

Write a function int fact(int n) which displays the factors of the integer n, and returns the number of factors. Call this function in main() with user input
#include<iostream>
using namespace std;
int fact(int n);
int main() {
int n,factor;
cout << "Enter an integer : ";
cin >> n;
factor = fact(n);
cout << factor;
return 0;
}
int fact(int n)
{
for (int i = 1; i <= n; ++i)
{
if (n % i == 0)
cout << i << endl;
}
return 0;
}
If I enter 7, I get 1,7,0 . How do i remove this 0 and how do i find the number of factors?

Answer 1

You should count in your int fact() function. Set a variable to 0 and increment each time you currently display i. Then at the end of the function instead of returning 0 return the count variable.
int fact(int n)
{
int count=0;
for (int i = 1; i <= n; ++i)
{
if (n % i == 0) {
cout << i << endl;
count++;
}
}
return count;
}

Answer 2

The key part is "and returns the number of factors". You don't do that. Keep a count of the factors:
int fact(int n)
{
int count = 0;
for (int i = 1; i <= n; ++i)
{
if (n % i == 0)
{
// found a factor, add to the count
count++;
cout << i << endl;
}
}
// return the count instead
return count;
}
Then, your main function can use that count:
factor = fact(n); // fact(n) will already print the factors
// now just print the number
cout << "Number of factors: " << factor << '\n';

Answer 3

#include <iostream>
#include <vector>
std::vector<int> fact(int n);
int main() {
int n;
std::cout << "Number: ";
std::cin >> n;
std::vector<int> factors = fact(n);
for (auto i : factors) {
std::cout << i << ' ';
}
std::cout << '\n';
std::cout << "Number of factors: " << factors.size() << '\n';
return 0;
}
std::vector<int> fact(int n) {
std::vector<int> vec{1};
for (int i = 2; i <= n / 2; ++i) {
if (n % i == 0) {
vec.push_back(i);
}
}
vec.push_back(n);
return vec;
}
If you're going to return anything from fact(), it should be the factors. To do so, I am using a std::vector. It is an array that can grow on demand. The numbers 1 and n are always factors, so I don't bother doing the math for them. The vector is initialized already holding the value 1, and I only calculate numbers up to and including half of n (Anything greater than n/2 won't divide evenly, so my loop is finished about half as fast by recognizing the actual range). I then just add n to the vector, which I return.
My main prints the vector, and the vector knows its own size, which is the number of factors.
Alternatively, you can just keep a count in your fact() function.
#include <iostream>
#include <vector>
// Prints factors of n and returns the number of factors
int fact(int n);
int main() {
int n;
std::cout << "Number: ";
std::cin >> n;
int numFactors = fact(n);
std::cout << "Number of factors: " << numFactors << '\n';
return 0;
}
int fact(int n) {
int factorCount = 2; // Already counting 1 and n
std::cout << "1 ";
for (int i = 2; i <= n / 2; ++i) {
if (n % i == 0) {
std::cout << i << ' ';
++factorCount;
}
}
std::cout << n << '\n';
return factorCount;
}

Answer 4

The main problem with your code is that your function always returns zero. You need to keep a count of factors and return it.
Besides that your code performance badly as the loop goes on much longer than needed. You can use the square root of n as the limit in the for loop. Like:
int fact(int n)
{
if (n < 1) return 0;
int res = 0;
int limit = sqrt(n);
for (int i = 1; i <= limit; ++i)
{
if (n % i == 0)
{
res += 2;
cout << i << " - " << n/i << endl;
}
}
if (limit * limit == n)
{
--res;
}
return res;
}
For n = 36 the output is:
1 - 36
2 - 18
3 - 12
4 - 9
6 - 6
and the returned value is 9
Below is another approach. It doesn't use square root. Instead it keeps the number of loops low by using the square of i as loop limit.
int fact(int n)
{
if (n < 1) return 0;
int res = 0;
int i = 1;
int i_square = i * i;
while (i_square < n)
{
if (n % i == 0)
{
res += 2;
cout << i << " - " << n/i << endl;
}
++i;
i_square = i * i;
}
if (i_square == n)
{
++res;
cout << i << " - " << n/i << endl;
}
return res;
}

Answer 5

Fact() always returns 0 so this line print 0
cout << factor;
for the number of factors you can change the return value of fact() :
int fact(int n)
{
int nb = 0;
for (int i = 1; i <= n; ++i)
{
if (n % i == 0) {
cout << i << endl;
nb++;
}
}
return nb;
}