Regex is not working in C

Question

I am using regex when I use it on shell it works but not inside the C program.

Any thoughts please?

echo "abc:[email protected]" | grep -E "(\babc\b|\bdef\b):[0-9]{10}@([A-Za-z0-9].*)"   //shell

reti = regcomp(&regex,"(\babc\b|\bdef\b):[0-9]{10}@([A-Za-z0-9].*)", 0); //c program

Answer 1

grep -E uses some enhanced ERE syntax meaning that the {n,m} quantifier braces (and also ( and )) do not have to be escaped (not the case in BRE regex).

You need to pass REG_EXTENDED flag to the regcomp, and also, since you can't use a word boundary, replace the first \b with (^|[^[:alnum:]_]) "equivalent". You need no trailing \b since there is a : in the pattern right after:

const char *str_regex = "(^|[^[:alnum:]_])(abc|def):[0-9]{10}@([A-Za-z0-9].*)";

The (^|[^[:alnum:]_]) part matches either the start of the string (^) or (|) a char other than alphanumeric or an underscore.

Full C demo:

#include <stdio.h>
#include <stdlib.h>
#include <regex.h>

int main (void)
{
  int match;
  int err;
  regex_t preg;
  regmatch_t pmatch[4];
  size_t nmatch = 4;
  const char *str_request = "abc:[email protected]";

  const char *str_regex = "(^|[^[:alnum:]_])(abc|def):[0-9]{10}@([A-Za-z0-9].*)";
  err = regcomp(&preg, str_regex, REG_EXTENDED);
  if (err == 0)
    {
      match = regexec(&preg, str_request, nmatch, pmatch, 0);
      nmatch = preg.re_nsub;
      regfree(&preg);
      if (match == 0)
        {
          printf("\"%.*s\"\n", pmatch[2].rm_eo - pmatch[2].rm_so, &str_request[pmatch[2].rm_so]);
          printf("\"%.*s\"\n", pmatch[3].rm_eo - pmatch[3].rm_so, &str_request[pmatch[3].rm_so]);
        }
      else if (match == REG_NOMATCH)
        {
          printf("unmatch\n");
        }
    }
  return 0;
 }