How do i continue this while - c++

Question

Im stuck with this program. What program does is to take a integer from user and display all number that are left after cutting the even numbers.
int main(){
long n, minder=0;
int cdonr, power=1;
cout<<"Give a positive number "<<endl;
cin>>n;
while (n>0){
cdonr=n%10;
if(cdonr % 2 != 0){
minder=minder+cdonr*power;
power=power*10;
}
n=n/10;
}
cout<<"The number that is left after all even number " << endl;
cout<<minder<<endl;
cout<<"Give a positive nr "<<endl;
cin>>n;
}
Can someone help me with this while because after it split the first numbers it gives no response at the second one.
Thanks to people on this community this is the answer
#include<iostream>
using namespace std;
int main() {
long n, minder = 0;
int cdonr, power = 1;
cout << "Give a positive number " << endl;
cin >> n;
while (n > 0) {
while (n > 0) {
cdonr = n % 10;
if (cdonr % 2 != 0) {
minder = minder + cdonr * power;
power = power * 10;
}
//always zero so we need a nested while loop
n = n / 10;
}
cout << "The number that is left after all even number " << endl;
cout << minder << endl;
//have to set values to default so they do not hold previous values
minder = 0;
cdonr = 0;
power = 1;
//we can reprompt user for a value
//n will get set and it validates with the outer while loop
//this allows it to run as many times as valid inputs
cout << "Give a positive number " << endl;
cin >> n;
}
}

Answer 1

Another answer as requested by user, exits upon == 0, <= 0, and not a number
#include<iostream>
using namespace std;
int main() {
long n, minder = 0;
int cdonr, power = 1;
cout << "Give a positive number " << endl;
cin >> n;
while (n > 0) {
while (n > 0) {
cdonr = n % 10;
if (cdonr % 2 != 0) {
minder = minder + cdonr * power;
power = power * 10;
}
//always zero so we need a nested while loop
n = n / 10;
}
cout << "The number that is left after all even number " << endl;
cout << minder << endl;
//have to set values to default so they do not hold previous values
minder = 0;
cdonr = 0;
power = 1;
//we can reprompt user for a value
//n will get set and it validates with the outer while loop
//this allows it to run as many times as valid inputs
cout << "Give a positive number " << endl;
cin >> n;
}
}

Answer 2

Try this, it is not infinite and you can do it as many times as you want by calling the function.
#include<iostream>
using namespace std;
void function(long n, long minder, int cdonr, int power) {
while (n > 0) {
cdonr = n % 10;
if (cdonr % 2 != 0) {
minder = minder + cdonr * power;
power = power * 10;
}
n = n / 10;
}
cout << "The number that is left after all even number " << endl;
cout << minder << endl;
}
int main() {
long n, minder = 0;
int cdonr, power = 1;
//can add a while loop here or just use a for
// if you know how many times you want
cout << "Give a positive number " << endl;
cin >> n;
if(n > 0){
function(n, minder, cdonr, power);
}
//probably add an else here in case of == 0 or < 0
// or just loop back if while
}

Answer 3

If you really want to do something forever, loop on it.
i.e. put your "function" to be repeated in a loop.
(And consider pulling it out to be an actual function)
int main() {
using namespace std;
while (true)
{
long n, minder = 0;
int cdonr, power = 1;
cout << "Give a positive number " << endl;
cin >> n;
while (n > 0) {
cdonr = n % 10;
if (cdonr % 2 != 0) {
minder = minder + cdonr*power;
power = power * 10;
}
n = n / 10;
}
cout << "The number that is left after all even number " << endl;
cout << minder << endl;
}
}