Best way to convert a regex match into an integer in C?

问题

I am using the standard <regex.h> library to match a complex string. Some matches are integers and my current solution is to use to_integer:

int to_integer(regmatch_t match, const char* str) {
    char buffer[16];
    int end = match.rm_eo - match.rm_so;
    memcpy(buffer, &str[match.rm_so], end);
    buffer[end] = '\0';
    return atoi(buffer);
}

struct data {
    int foo, bar, baz;
};

struct data fetch(char *string) {
   int groups = 4;
   regmatch_t matches[groups];
   if (regexec(&regex, string, groups, matches, 0)) abort();

   return (struct data){
       to_integer(matches[1], string),
       to_integer(matches[2], string),
       to_integer(matches[3], string)
   };
}

Is there a more elegant way that does not involve an intermediate buffer?

Without an intermediate buffer, the following would eventually fail: ([0-9]{3})[0-9]{2}. I also cannot modify str in place because it is constant.

EDIT

From this question I wrote the following:

int to_integer(regmatch_t match, const char* str) {
    const char *end = &str[match.rm_eo];
    return strtol(&str[match.rm_so], (char**)&end , 10);
}

Unfortunately, the explicit cast (char*) is quite ugly. My previous solution involving a copy of the string looks a bit safer IMHO.

来源：https://stackoverflow.com/questions/64412234/best-way-to-convert-a-regex-match-into-an-integer-in-c