Thread: Roman numeral converter problem

First off I am quite new to programing so the solution is probably quite obvious (ether that or I wrote this program horribly wrong), so bear with me here. I am trying to make a program that converts Roman numerals to Arabic ones and it isn't doing what I thought it would. When I input, for example, "VII" it comes back with something like -2 million!?! BTW the comments are what I think the program does, not necessarily what it actually does. Thanks in advance.

Code:

#include <stdio.h>		/*BEWARE!! THIS DOES NOT WORK YET*/
#include <string.h>
#include <stdlib.h>

void stoupper(char *s) /*definition for string upercase conversion*/
{
for(; *s; s++)
if(('a' <= *s) && (*s <= 'z'))
*s = 'A' + (*s - 'a');
}

char rome[20];
int arab=0;
int lngth;
int count;

int main()
{
	fgets(rome, 20, stdin);		/*gets the roman numeral*/
	stoupper(rome);		/*puts roman numeral in uppercase*/
	lngth=strlen(rome);		/*finds length of numeral*/
	printf("You entered %s\n",rome);
	printf("that is %d charicters long.\n", lngth-1);
	int buf[lngth];		/*for adding up numbers that equal eachother*/
	int lyr[lngth];		/*for the final tally*/
	for(count=0; count<lngth; count++)	/*puts a number value to each letter*/
	{
		switch(rome[count])
		{
			case 'M':
				buf[count]=1000;
				break;
			case 'D':
				buf[count]=500;
				break;
			case 'C':
				buf[count]=100;
				break;
			case 'L':
				buf[count]=50;
				break;
			case 'X':
				buf[count]=10;
				break;
			case 'V':
				buf[count]=5;
				break;
			case 'I':
				buf[count]=1;
				break;
			default:
			
				break;
		}
	}
	int t=0;
	for(count=0; count<lngth; count++)		/* adds up numbers that are the same and moves them to a new aray*/
	{
		if(buf[count]==buf[count+1]) 
		{
			lyr[t]=buf[count]*2;
			t++;
			count++;
		}
		if(buf[count]==buf[count+1] && buf[count]==buf[count+2])
		{
			lyr[t]=buf[count]*3;
			t++;
			count=count+2;
		}
		else
		{
			lyr[t]=buf[count];
			t++;
		}
	}
	for(count=0; count<=t; count++)		/* finds groups that subtract from the total */
	{
		if(lyr[count]<lyr[count+1])
		{
			lyr[count]=lyr[count]*-1;
		}
	}
	for(count=0; count<t; count++)		/*gets final sum */
	{
		arab=arab+lyr[count];
	}
	printf("In arabic:-%d", arab);
}

Originally Posted by 7heTexanRebel

My goal is to convert roman numerals to arabic(pretty sure they're the ones we use ie. 123..etc.) So I'm trying to convert VII to 7. I'm guessing that the overflow is when it tries to read "count+1" or "count+2" when it is at the end of the array and there is no "count+1 or 2"? Sorry for my ignorance. I'm not sure I see the "adds up numbers that are the same and moves them to a new array" problems. It's supposed to see, for example, in VII that "II" are both 1 so "II" is 2, so it treats "VII" as 5 and 2 instead of 5, 1, and 1. (supposed to make them easier to add up)

MK27,
Thanks! It looks like I was over complicating the program. =\

Our numbers are indeed arabic numerals. The Greeks (and others) extended geometry significantly, the Arab scholars did the same with algebra, including our number representations.

But that won't help your program.

I like the switch statement - well done on that. Your programs lacks any logic to handle the subtraction needed when a smaller roman numeral, is directly to the left of, a larger roman numeral: e.g.: IV, IX, XD, etc.

As MK27 pointed out, you only need to check ONE numeral to the right of each digit, to see if the subtraction should be done. The amount you should subtract is the lower (first if you read it from left to right) numeral: e.g. IX, adds up to "11", but you now need to subtract 1 from that, because the I is immediately to the left of the larger valued X. Bringing your total to 10 - but you also (wrongly), have added I to the X initially, which needs to be corrected, so subtract one more I value, bringing the total to 9.

So IX = 11 - (2*I)