Curley braces are not working in regular expression in C language - c

Question

The Curly braces{} are not working in C language Regular expressions, it is always giving output as NO match , if i give correct input as "ab" or "ac". I would request to help in this case.
#include <sys/types.h>
#include <regex.h>
#include <stdio.h>
int main(int argc, char *argv[]){ regex_t regex;
int reti;
char msgbuf[100];
/* Compile regular expression */
reti = regcomp(&regex, "[a-c]{2}", 0);
if( reti ){ fprintf(stderr, "Could not compile regex\n"); return(1); }
/* Execute regular expression */
reti = regexec(&regex, "ab", 0, NULL, 0);
if( !reti ){
puts("Match");
}
else if( reti == REG_NOMATCH ){
puts("No match");
}
else{
regerror(reti, &regex, msgbuf, sizeof(msgbuf));
fprintf(stderr, "Regex match failed: %s\n", msgbuf);
return 1;
}
/* Free compiled regular expression if you want to use the regex_t again */
regfree(&regex);
return 0;
}

Answer 1

You are using the Basic Regular Expressions dialect that has no knowledge of the quantifier {n} in the regex.
One solution would be to supply the option REG_EXTENDED as the last argument instead of 0 when creating your regex_t object.
reti = regcomp(&regex, "[a-c]{2}", REG_EXTENDED);
See http://ideone.com/oIBXxu for a Demo of your code with my modification.
As Casimir et Hippolyte notes in the comments Basic Regular Expressions support the {} quantifier as well but the curly braces must be escaped with a \ in the regex which again has to be escaped in the C string as \\. So you can use the line
reti = regcomp(&regex, "[a-c]\\{2\\}", 0);
as well as an alternative to the solution above(running Demo with this line modified under http://ideone.com/x7vlIO).
You can check http://www.regular-expressions.info/posix.html for more information about the difference between Basic and Extended Regular Expressions.