How to convert integer value to Roman numeral string?

Question

How can I convert an integer to its String representation in Roman numerals in C ?

Answer 1

The easiest way is probably to set up three arrays for the complex cases and use a simple function like:

// convertToRoman:
//   In:  val: value to convert.
//        res: buffer to hold result.
//   Out: n/a
//   Cav: caller responsible for buffer size.

void convertToRoman (unsigned int val, char *res) {
    char *huns[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
    char *tens[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
    char *ones[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
    int   size[] = { 0,   1,    2,     3,    2,   1,    2,     3,      4,    2};

    //  Add 'M' until we drop below 1000.

    while (val >= 1000) {
        *res++ = 'M';
        val -= 1000;
    }

    // Add each of the correct elements, adjusting as we go.

    strcpy (res, huns[val/100]); res += size[val/100]; val = val % 100;
    strcpy (res, tens[val/10]);  res += size[val/10];  val = val % 10;
    strcpy (res, ones[val]);     res += size[val];

    // Finish string off.

    *res = '\0';
}

This will handle any unsigned integer although large numbers will have an awful lot of M characters at the front and the caller has to ensure their buffer is large enough.

Once the number has been reduced below 1000, it's a simple 3-table lookup, one each for the hundreds, tens and units. For example, take the case where val is 314.

val/100 will be 3 in that case so the huns array lookup will give CCC, then val = val % 100 gives you 14 for the tens lookup.

Then val/10 will be 1 in that case so the tens array lookup will give X, then val = val % 10 gives you 4 for the ones lookup.

Then val will be 4 in that case so the ones array lookup will give IV.

That gives you CCCXIV for 314.

---

A buffer-overflow-checking version is a simple step up from there:

// convertToRoman:
//   In:  val: value to convert.
//        res: buffer to hold result.
//   Out: returns 0 if not enough space, else 1.
//   Cav: n/a

int convertToRoman (unsigned int val, char *res, size_t sz) {
    char *huns[] = {"", "C", "CC", "CCC", "CD", "D", "DC", "DCC", "DCCC", "CM"};
    char *tens[] = {"", "X", "XX", "XXX", "XL", "L", "LX", "LXX", "LXXX", "XC"};
    char *ones[] = {"", "I", "II", "III", "IV", "V", "VI", "VII", "VIII", "IX"};
    int   size[] = { 0,   1,    2,     3,    2,   1,    2,     3,      4,    2};

    //  Add 'M' until we drop below 1000.

    while (val >= 1000) {
        if (sz-- < 1) return 0;
        *res++ = 'M';
        val -= 1000;
    }

    // Add each of the correct elements, adjusting as we go.

    if (sz < size[val/100]) return 0;
    sz -= size[val/100];
    strcpy (res, huns[val/100]);
    res += size[val/100];
    val = val % 100;

    if (sz < size[val/10]) return 0;
    sz -= size[val/10];
    strcpy (res, tens[val/10]);
    res += size[val/10];
    val = val % 10;

    if (sz < size[val) return 0;
    sz -= size[val];
    strcpy (res, ones[val]);
    res += size[val];

    // Finish string off.

    if (sz < 1) return 0;
    *res = '\0';
    return 1;
}

although, at that point, you could think of refactoring the processing of hundreds, tens and units into a separate function since they're so similar. I'll leave that as an extra exercise.

Answer 2

don't use a pre-calculated map for the troublesome bits! this is a solved problem in compilers (grammar rules).

/* roman.c */
#include <stdio.h>

/* LL(1) roman numeral conversion */
int RN_LL1 (char *buf, const size_t maxlen, int n)
{
  int S[]  = {    0,   2,   4,   2,   4,   2,   4 };
  int D[]  = { 1000, 500, 100,  50,  10,   5,   1 };
  char C[] = {  'M', 'D', 'C', 'L', 'X', 'V', 'I' };
  const size_t L = sizeof(D) / sizeof(int) - 1;
  size_t k = 0; /* index into output buffer */
  int i = 0; /* index into maps */
  int r, r2;

  while (n > 0) {
    if (D[i] <= n) {
      r = n / D[i];
      n = n - (r * D[i]);
      /* lookahead */
      r2 = n / D[i+1];
      if (i < L && r2 >= S[i+1]) {
        /* will violate repeat boundary on next pass */
        n = n - (r2 * D[i+1]);
        if (k < maxlen) buf[k++] = C[i+1];
        if (k < maxlen) buf[k++] = C[i-1];
      }
      else if (S[i] && r >= S[i]) {
        /* violated repeat boundary on this pass */
        if (k < maxlen) buf[k++] = C[i];
        if (k < maxlen) buf[k++] = C[i-1];
      }
      else
        while (r-- > 0 && k < maxlen)
          buf[k++] = C[i];
    }
    i++;
  }
  if (k < maxlen) buf[k] = '\0';
  return k;
}

/* gcc -Wall -ansi roman.c */
int main (int argc, char **argv)
{
  char buf[1024] = {'\0'};
  size_t len;
  int k;
  for (k = 1991; k < 2047; k++)
  {
    len = RN_LL1(buf, 1023, k);
    printf("%3lu % 4d %s\n", len, k, buf);
  }
  return 0;
}

you don't actually need to declare S either. it should be easy to see why after tracing a few examples, but, the sequence will just repeat indefinitely over (i % 2) with a special case for zero.

Answer 3

    static string ConvertToRoman(int num)
    {
        int d = 0;
        string result = "";
        while (num > 0)
        {
            int n = num % 10;
            result = DigitToRoman(n, d) + result;
            d++;
            num = num / 10;
        }
        return result;
    }
    static string DigitToRoman(int n, int d)
    {
        string[,] map = new string[3, 3] { { "I", "V", "X" }, { "X", "L", "C" }, { "C", "D", "M" } };
        string result="";
        if (d <= 2)
        {
            switch (n)
            {
                case 0:
                    result = "";
                    break;
                case 1:
                    result = map[d, 0];
                    break;
                case 2:
                    result = map[d, 0] + map[d, 0];
                    break;
                case 3:
                    result = map[d, 0] + map[d, 0] + map[d, 0];
                    break;
                case 4:
                    result = map[d, 0] + map[d, 1];
                    break;
                case 5:
                    result = map[d, 1];
                    break;
                case 6:
                    result = map[d, 1] + map[d, 0];
                    break;
                case 7:
                    result = map[d, 1] + map[d, 0] + map[d, 0];
                    break;
                case 8:
                    result = map[d, 1] + map[d, 0] + map[d, 0] + map[d, 0];
                    break;
                case 9:
                    result = map[d, 0] + map[d, 2];
                    break;
            }
        }
        else if (d == 3 && n < 5)
        {
            while (--n >= 0)
            {
                result += "M";
            }
        }
        else
        {
            return "Error! Can't convert numbers larger than 4999.";
        }
        return result;
    }