12
\$\begingroup\$

I've created a regular expression (regex) parsing library in C, and would like some feedback on it. Speed is really important to me, but any and all suggestions are acceptable.

#include <ctype.h>

static int regex_matchHere(const char *regex, char *s, int *len);
static int regex_matchGroup(int c, int group);
static int regex_matchQuantity(int quant, int c, const char *regex, char *s, int *len);

int regex_match(const char *regex, char *s, int *len)
{
    char *p = s;

    /* force match from the beginning of the string */
    if (regex[0] == '^') return (regex_matchHere(regex + 1, s, len) ? 0 : -1);

    /* iterate the string to find matching position */
    do
    {
        *len = 0;
        if (regex_matchHere(regex, p, len)) return (int)(p - s);
    } while (*p++ != '\0');
    return -1;
}

static int regex_matchHere(const char *regex, char *s, int *len)
{
    int c = regex[0];

    if (regex[0] == '\0') return 1; /* end of regex = full match */
    else if (regex[0] == '$' && regex[1] == '\0') return (*s == '\0'); /* check end of string */
    else if (regex[0] == '\\' && regex[1] != '\0') /* check escaped symbol */
    {
        c = regex[1];
        if (c != '^' && c != '$' && c != '\\' && c != '+' && c != '*' && c != '-' && c != '?') c = c | 0x100;
        regex = regex + 1;
    }
    /* check for special operators *,+,?,- */
    if (regex[1] == '*' || regex[1] == '+' || regex[1] == '-' || regex[1] == '?') return regex_matchQuantity(regex[1], c, regex+2, s, len);
    else if (*s != '\0' && regex_matchGroup(*s, c))
    {
        *len = *len + 1;
        return regex_matchHere(regex+1, s+1, len);
    }
    return 0;
}

static int regex_matchGroup(int c, int group)
{
    if ((group & 0xff) == '.') group ^= 0x100;
    if (group < 0x100) return c == group; /* a single char */
    /* a meta char, like \d, ... */
    switch (group & 0xff)
    {
        case 'd': return isdigit(c);
        case 's': return isspace(c);
        case 'D': return !isdigit(c);
        case 'S': return !isspace(c);
        case '.': return 1;
    }
    return 0;
}

static int regex_matchQuantity(int quant, int c, const char *regex, char *s, int *len)
{
    if (quant == '?')
    {
        if (regex_matchGroup(*s, c))
        {
            *len = *len + 1;
            s = s + 1;
        }
        return regex_matchHere(regex, s, len);
    }

    if (quant == '+' || quant == '*') /* match as much as possible */
    {
        char *p;
        for (p = s; *p != '\0' && regex_matchGroup(*p, c); p++) *len = *len + 1;
        if (quant == '+' && p == s) return 0;
        do
        {
            if (regex_matchHere(regex, p, len)) return 1;
            *len = *len - 1;
        } while (p-- > s);
    }
    else if (quant == '-') /* match as little as possible */
    {
        do
        {
            if (regex_matchHere(regex, s, len)) return 1;
            *len = *len + 1;
        } while (*s != '\0' && regex_matchGroup(*s++, c));
    }
    return 0;
}
\$\endgroup\$
1
  • 2
    \$\begingroup\$ Although it can be divined from the code, it wouldn't hurt to explicitly state the exact sort of regular expressions this is intended to parse/match. \$\endgroup\$ Mar 11, 2014 at 14:50

1 Answer 1

9
\$\begingroup\$

What you did well

The code seems clean and logically organized. I like your 0x100-bit hack to indicate special characters. You could make that convention more obvious in the comments, though.

What you could improve on

  1. The return value of regex_match() is weird. I'd like it to return a non-zero value if the match succeeded, and a zero value if the match failed, so that I can call it like this:

    if (regex_match(...)) {
        // Do stuff for successful match
    } else {
        // Do stuff for failed match
    }
    

    Trying to return the position of the match just leads to confusion, reminiscent of the way PHP's strpos() returns 0 to indicate a successful match at the beginning of the subject (but FALSE to indicate a non-match). You don't want to be like PHP, do you?

    I suggest that the signature for regex_match() should look like this:

    /**
     * Returns 1 if matched, 0 if not matched.
     *
     * Pass a pointer to a match_result if you care to find out the
     * details of the match (its length, position, and possibly other
     * information supported in the future, such as parenthesized
     * capture groups), or pass a NULL if you don't care about the details.
     */
    int regex_match(const char *regex, const char *subject, struct match_result *result);
    

    Alternatively, return a pointer to a new struct match_result if the match succeeded. The caller would have to free() the result later, though, so I don't like it as much.

  2. Regular expressions often include modifier flags, such as a case-insensitive flag or a continue-searching-where-the-previous-match-ended flag. You might want to plan your interfaces accordingly. (To support the latter, the struct match_result* would probably become an in-out parameter rather than an out-parameter.)

  3. For performance, regular expressions are frequently compiled into an automaton. You interpret the regular expressions as you go. You may wish design the library's interface to have a regex_compile() function that transforms the expression into a struct that is meaningful to your library but opaque to the user. For now, the "compilation" could just be the identity transformation; you can enhance it later when the need for better performance arises or when you enhance the feature set of the regular expressions.

  4. The function name regex_matchGroup() confuses me. "Group" implies something like parentheses, I think. regex_matchAtom() might be a more appropriate name.

  5. You need unit tests!

\$\endgroup\$
1
  • 4
    \$\begingroup\$ This is a nice review, except for the part about the return value – returning an index an expecting the caller to do if (0 <= regex_match(...)) is perfectly natural. The problem with PHP is that it took a C-ish idiom and messed it up by returning false instead of -1, but that does not discredit the original idiom. \$\endgroup\$
    – amon
    Mar 11, 2014 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.