C++ Compute prime factors of a number and print them - c++

Question

I want to ask the user for an integer, and print the prime factors.
Example: User enters 100, program displays 2 2 5 5
So far I have the following:
#include <iostream>
using namespace std;
void factors(int n){
int z = 2;
while (z * z <= n)
{
if (n % z == 0)
{
cout << z;
n = (n / z);
}
else
{
z++;
}
}
if (n > 1)
{cout << n << "\n";}
}
int main()
{
int x;
cout << "Input positive integer greater than 1: ";
cin >> x;
factors(x);
cout << "The result: " << x;
return 0;}
My question is how do I get my main function to communicate with the factors procedure. I run the program, I get the message asking for input, I input 12, I get the message "The result" but with a number of 25, and also 12, the number that the user input. it's like the program is avoiding my factors(int n) procedure. Help with the syntax please?!?
My issue is with the syntax I think.
Because I found the following function to help with listing prime factors:
Finding prime factors
-user44810
define factors(n)
z = 2
while (z * z <= n)
if (n % z == 0)
output z
n /= z
else
z++
if n > 1
output n
Thank you!!!

Answer 1

Try this
#include <iostream>
#include <vector>
using namespace std;
vector<int> factors(int n){
vector<int> result;
int z = 2;
while (z * z <= n)
{
if (n % z == 0)
{
result.push_back(z);
n = (n / z);
}
else
{
z++;
}
}
if (n > 1)
result.push_back(z);
return result;
}
int main()
{
int x;
cout << "Input positive integer greater than 1: ";
cin >> x;
vector<int> result_factors = factors(x);
cout << "The result: ";
for (int i: result_factors)
cout << "i ";
cout << endl;
return 0;
}
I changed your factor() function to output nothing on cout but saving the factors in a vector and returning it. In the main function, iterate over the result vector and print the values to cout.

Answer 2

So the biggest issue is that there is a missing brace at the end of your factors function. you need to add another brace after the if (n > 1) brace. Also, there is a missing semicolon at the end of the last cout that will throw an error.
Another issue that won't prevent the code from running is that when you print "The result: " << x you will be giving the same value as the user input.
If you want your main function to print out the result from factors() at that spot, then the function needs to return the data instead of printing it. To fix this, the factors function can be made to return a string with the output you want:
//return a string with the output
string factors(int n){
//create a string to save the output to
string factorlist = "";
int z = 2;
while (z * z <= n)
{
if (n % z == 0)
{
//append z to the end of the string and add a space to make it easier to read
factorlist+=to_string(z)+" ";
n = (n / z);
}
else
{
z++;
}
}
if (n > 1)
{
//append n to the end of the string and add a newline
factorlist+=to_string(n)+"\n";
}
//output the string factorlist to wherever the function was called from
return factorlist;
}
then on the lines that look like:
factors(x);
cout << "The result: " << x
Should be:
cout << "The result: " << factors(x);
Currently, you are just printing out the value of x that the user input. If you want to save the value of factors(x) you would need to set a variable equal to it like this:
string FactorsResult = factors(x)
And then print it; or, as in the corrected code above, just print it directly in the cout statement.