Wednesday, November 18, 2009

regexp.c - Regular Expression Sample Example

regexp.c - Regular Expression Sample Example

/*******************************************************
Copyright GPL Mike Chirico mmc mchirico@users.sourceforge.net
http://sourceforge.net/project/showfiles.php?group_id=79066



Note the possible combinations with '(1|2)(a|b)'.

     The complete match '1b'
     The match with '1'
     The match with 'b'

./regexp '(1|2)(a|b)' 1b
match
2 0
$0 = 1b, preg.re_nsub = 2
1 0
$1 = 1, preg.re_nsub = 2
2 1
$2 = b, preg.re_nsub = 2




 ./regexp '([a|c|e|g]{2}|[h-z])([0-9]|-)(a|b)' cc8b
match
4 0
$0 = cc8b, preg.re_nsub = 3
2 0
$1 = cc, preg.re_nsub = 3
3 2
$2 = 8, preg.re_nsub = 3
4 3
$3 = b, preg.re_nsub = 3


*******************************************************/
#include <stdio.h>
#include <sys/types.h>
#include <string.h>
#include <regex.h>

#define NUM_MATCHES 4//max sub-matches

int
main (int argc, char *argv[])
{
  regex_t preg;
  regmatch_t pmatch[NUM_MATCHES];
  size_t rm, i;
  char pom[1024];


  if (argc != 3)
    {
      printf ("Usage ./regexp [a-z]{2}[0-9]{2} a2b3ed23\n\n"
              " this program is setup so that there will be a maximum\n"
              " of 4 submatches\n"
              "        ./regexp '([a|c|e|g]{2}|[h-z])([0-9]|-)(a|b)' cc8b \n"
             "        match\n"
             "        4 0\n"
             "        $0 = cc8b, preg.re_nsub = 3\n"
             "        2 0\n"
             "        $1 = cc, preg.re_nsub = 3\n"
             "        3 2\n"
             "        $2 = 8, preg.re_nsub = 3\n"
             "        4 3\n"
       "        $3 = b, preg.re_nsub = 3\n");

      return 1;
    }

  /* it's possible something won't compile like  ./regexp '*' abc */
  if ((rm = regcomp (&preg, argv[1], REG_EXTENDED)) != 0)  
    {
      fprintf(stderr,"Invalid expression:'%s'\n",argv[1]);
      return 1;
    }
  (rm =
   regexec (&preg, argv[2], NUM_MATCHES, pmatch,
     0)) ? printf ("No match\n") : printf ("match\n");
  for (i = 0; !rm && i <= preg.re_nsub; i++)
    {
      strncpy (pom, argv[2] + pmatch[i].rm_so,
        pmatch[i].rm_eo - pmatch[i].rm_so);
      printf ("%d %d\n", pmatch[i].rm_eo, pmatch[i].rm_so);
      pom[pmatch[i].rm_eo - pmatch[i].rm_so] = '\0';
      printf ("$%d = %s, preg.re_nsub = %d\n", i, pom, preg.re_nsub);
    }

  regfree (&preg);

  return 0;
}


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