C solution to Roman2Int problem

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    It is obvious that Roman numerals consist of 'I', 'V', 'X', 'L', 'C', 'D', 'M', representing for 1, 5, 10, 50, 100, 500, 1000. We also know that a number reside to the left of a larger number means subtraction operation. Symmetrically, a number reside to the right of an equal or larger number means addition operation. With that in mind, let's look at 2 examples:

    1. IV
      The answer is 4, because 'I' is smaller than as well as left to 'V', so we interpret it as: 'V' - 'I' = 4.
    2. LXV
      The answer is 65, because 'X' is larger than and right to 'V', so we add them: 'X' + 'V' = 15; meanwhile, 'L' and 'X' show us the same pattern, we we do addition again: 'L' + previous result = 65.

    Now we have no trouble interpreting Roman numerals. In my solution, I firstly define a function for the convenience of transferring a Roman char to an int type, then apply algorithm on it:

    int charToInt( char c ) {
        switch( c ) {
            case 'I':
                return 1;
            case 'V':
                return 5;
            case 'X':
                return 10;
            case 'L':
                return 50;
            case 'C':
                return 100;
            case 'D':
                return 500;
            case 'M':
                return 1000;
                return -1;
    int romanToInt(char* s) {
        char prev, curr;
        int sum, len, a, b;
        len = strlen( s );
        prev = s[len - 1];
        sum = charToInt( prev );
        while( --len > 0 ) {
            curr = s[len - 1];
            a = charToInt( prev );
            b = charToInt( curr ) ;
            prev = curr;
            sum = a > b ? sum - b : sum + b;
        return sum;

    However, my solution does not check error for illegal inputs that violate the rules of representing Roman numerals. Solution rests on the surmise of correct input.

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