Reversing a number C++ - c++

Question

Looking for some advice here on what I'm getting wrong. Everything in my main should be fine and left unchanged. My problem is in my reverse function. It's printing the reversed number right before the cout statement of "The number is" instead down below where it should be. I spent awhile trying to fix but can't come up with a solution.
#include <iostream>
#include <iomanip>
#include <cstdlib>
#include <ctime>
using namespace std;
const int NUM_VALS = 10; //the maximum number of values to use
int reverse(int num);
bool isPrime(int num);
int main()
{
int number, //Holds the random number that is manipulated and tested
loopCnt; //Controls the loop
//set the seed value for the random number generator
//Note: a value of 1 will generate the same sequence of "random" numbers every
// time the program is executed
srand(1);
//Generate 10 random numbers to be manipulated and tested
for( loopCnt = 1; loopCnt <= NUM_VALS; loopCnt++ )
{
//Get a random number
number = rand();
//Display the sum of adding up the digits in the random number, the reversed
//random number, and whether or not the number is palindromic or a prime number
cout << "The number is " << number << endl
<< "----------------------------------------" << endl
// << "Adding the digits result" << setw(16) << sumDigits( number ) << endl
<< "Reversing the digits result" << setw(13) << reverse(number) << endl
// << "Is the number a palindrome?" << setw(13) << (isPalindrome(number)? "Yes" : "No") << endl
// << "Is the number prime?" << setw(20) << (isPrime(number)? "Yes" : "No") << endl
<< endl << endl;
}
return 0;
}
int reverse(int num)
{
int quo, rem;
quo = num;
while (quo != 0)
{
rem = quo % 10;
cout << rem;
quo /= 10;
}
}
bool isPrime(int num)
{
int i;
if (num % 2 == 0)
return false;
for (i = 3; i*i <= num; i+=2)
{
if (num % i == 0)
return false;
}
return true;
}

Answer 1

You need to have your reverse function return the number as reversed, because the return value is used in main.
You can build the reversed number by multiplying a "reversed" value by 10, then adding in the remainder:
int reverse(int num)
{
int reversed = 0;
int quo, rem;
quo = num;
while (quo != 0)
{
rem = quo % 10;
reversed = reversed * 10 + rem;
quo /= 10;
}
return reversed;
}

Answer 2

You can also use this method to reverse a number by taking string input and then reverse it and convert it to int.
#include <iostream>
#include<string>
using namespace std;
int reverse_num(string a)
{
string s;
for(int i=a.length()-1;i>=0;i--)
{
s+=a[i];
}
int n;
n=stoi(s);
return n;
}
int main()
{
string a;
cin>> a;
cout<<reverse_num(a);
return 0;
}