I'm just looking for a hint in the right direction if someone could help me please. The first part works good, but i can't the the call by reference to reverse the digits for me. It outputs a 0 or it outputs the numbers in the same order that i input them.

Code:// reversedigits.cpp // written by // cis #include <iostream> #include <iomanip> using namespace std; int revdigits ( int ); // call by value void revdigits1 ( int, int& );// call by reference int width ( int ) ; void width1 ( int, int& ); int main() { int number = 0, r = 0; cout << "Enter a number between 1 and 9999: " << fixed; cin >> number; cout << "The number with its digits reversed is: " << setw ( width ( number ) ) << revdigits ( number ) << endl; void revdigits1 ( int n, int r ); cout << "The number with its digits reversed is: " ; void width1 ( int n,int r ); cout << r << endl; return 0; } int revdigits ( int n ) { int reverse = 0, divisor = 1000, multiplier = 1; while ( n > 10 ) { if ( n >= divisor ) { reverse += n / divisor * multiplier; n %= divisor; divisor /= 10; multiplier *= 10; } else divisor /= 10; } reverse += n * multiplier; return reverse; } void revdigits1 ( int n, int &r ) { int reverse = 0, divisor = 1000, multiplier = 1; while ( n > 10 ) { if ( n >= divisor ) { reverse += n / divisor * multiplier; n %= divisor; divisor /= 10; multiplier *= 10; } else divisor /= 10; } reverse += n * multiplier; r = reverse; } int width ( int n ) { if ( n / 1000 ) return 4; else if ( n / 100 ) return 3; else if ( n / 10 ) return 2; else return 1; } void width1 ( int n, int &r ) { if ( n / 1000 ) r = 4; else if ( n / 100 ) r = 3; else if ( n / 10 ) r = 2; else r = 1; }