Converting int to roman numerals, only end if zero entered [duplicate] - c++

Question

This question already has answers here:
What is an undefined reference/unresolved external symbol error and how do I fix it?
(39 answers)
Closed 6 years ago.
I need to convert int to Roman numerals, my code is failing to compile with mingw using
C:\code>g++ -Wall -o rome_to_int rome_to_int.cpp
C:\Users\vm1\AppData\Local\Temp\ccSvDFll.o:task2.cpp:(.text+0x1278): undefined r
eference to `get_int_ref_checked()'
collect2.exe: error: ld returned 1 exit status
Now as part of the code, it must keep asking for input, converting that input untill zero is entered. Once Zero has been entered then it can exit.
I must use code from a privius question which covers the continuing untill which is found in the get_int_ref_checked, I can assume ONLY integers will be entered and nothing else, no doubles nor string of words.
I'm not sure what else to check.
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int get_int_ref_checked();
string int_to_ROMAN();
/**
Turns the number from 1 to 9
*/
string digit_name(int digit)
{
if (digit == 1) return "I";
if (digit == 2) return "II";
if (digit == 3) return "III";
if (digit == 4) return "IV";
if (digit == 5) return "V";
if (digit == 6) return "VI";
if (digit == 7) return "VII";
if (digit == 8) return "VIII";
if (digit == 9) return "IX";
return "";
}
/**
Turns the number 10, 20, 30, 40, 50, 60, 70, 80, 90
*/
string ten_name(int n)
{
if (n == 1) return "X";
if (n == 2) return "XX";
if (n == 3) return "XXX";
if (n == 4) return "XL";
if (n == 5) return "L";
if (n == 6) return "LX";
if (n == 7) return "LXX";
if (n == 8) return "LXXX";
if (n == 9) return "XC";
return "";
}
/**
Turns the number 100, 200, 300, 400, 500, 600, 700, 800, 900
*/
string hun_name(int n)
{
if (n == 1) return "C";
if (n == 2) return "CC";
if (n == 3) return "CCC";
if (n == 4) return "CD";
if (n == 5) return "D";
if (n == 6) return "DC";
if (n == 7) return "DCC";
if (n == 8) return "DCCC";
if (n == 9) return "CM";
return "";
}
/**
Turns the number 1000, 2000, 3000
*/
string thou_name(int n)
{
if (n == 1) return "M";
if (n == 2) return "MM";
if (n == 3) return "MMM";
return "";
}
string int_to_ROMAN(int n)
{
int part = n; // The part that still needs to be converted
string roman; // The return value
if (part >= 1000)
{
roman = thou_name(part / 1000);
part = part % 1000;
}
if (part >= 100)
{
roman = roman + hun_name(part/100);
part = part % 100;
}
if (part >= 10)
{
roman = roman + ten_name(part/10);
part = part % 10;
}
if (part > 0)
{
roman = roman + digit_name(part);
}
return roman;
}
int main()
{
int n;
string r;
n = get_int_ref_checked();
if (n == 0)
{
cout<< "The intput is ZERO! ";
}
if(n > 0)
{
cout<< "The convert is " << int_to_ROMAN(n);
}
return 0;
}
void get_int_ref_checked(int& n) {
for(;;) // infinite loop
{
cout<< "Please enter an integer\n";
std::string inp;
std::stringstream ss;
ss << inp;
if( !(ss >> n) || n >= 4000 || n <= 0)
{
cout << "Error enter a number n: 0 < n < 4000: \n";
}
}
}

Answer 1

You are forward declaring the function int get_int_ref_checked() but there is no implementation, as int get_int_ref_checked(int& n) is another function, which is not being forward declared and so results in the error your seeing.