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.

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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 sissy pre-calculated map for the difficult cases.

/* roman.c */ #include <stdio.h>  /* LH(1) roman numeral conversion */ int RN_LH1 (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_LH1(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.