C语言中的Posix正则表达式

Question

I am working on this code where i have to compile some regular expression and use these compiled versions multiple times on different strings. So i decided to make a function where i could pass these compiled version for matching the string. My problem is that when i pass the compiled version in the function its showing a match but setting the regmatch_t structure fields to 0. However if i use them within the same function i am getting correct results.

void match_a(regex_t *a,char *str)
{
  regmatch_t match_ptr;
  size_t nmatch;
  regexec(a,str,nmatch,&match_ptr,0);
}
int main()
{
  regex_t a;
  regmatch_t match_ptr;
  size_t nmatch;
  char *str="acbdsfs";
  regcomp(&a,str,RE_EXTENDED);
  match_a(&a,str);
}

This is the general structure of the code.Please suggests any ways to debug this program

Answer 1

I'm not sure you understand how to use regexec . The nmatch argument tells regexec the number of regmatch_t objects you have provided. You haven't initialised the nmatch variable so it could be any indeterminate value, which will likely lead to a crash at some stage, or it may be 0 in which case the regexec function is defined to ignore the pmatch argument.

If you want only one regmatch_t result, try this:

void match_a(regex_t *a,char *str)
{
    regmatch_t match;
    size_t nmatch = 1;

    regexec(a, str, nmatch, &match, 0);
}

If you want up to 10 regmatch_t (for regular expressions with groups etc), try this:

void match_a(regex_t *a,char *str)
{
    regmatch_t matches[10];
    size_t nmatch = 10;

    regexec(a, str, nmatch, matches, 0);
}

For more information, read this documentation .

Answer 2

Why matches is not filled in position 1?

    regex_t a;
    regcomp(&a,"brasil",REG_ICASE);

    regmatch_t matches[2];
    size_t nmatch = 2;
    regexec(&a,"brasil brasil",nmatch,matches,0);

    int x;
    for(x=0;x<2;x++)
            printf("%i\n",matches[x].rm_so);