I have witten a simple c program as follows:
#include <stdio.h>
#include <regex.h>
int main(int argc, char* argv[])
{
regex_t re;
char *pattern = "\\d{3}([a-zA-Z]+).(\\d{2}|N/A)\\s\\1"; // use \d, but wouldn't match
const char *target = "123Hello N/A Hello";
regcomp(&re, pattern, REG_EXTENDED|REG_NOSUB);
int ret = regexec(&re, target, (size_t) 0, NULL, 0);
if(ret == REG_NOMATCH) {
printf("\n%s\n does not match \n%s\n", target, pattern);
}
pattern = "[0-9]{3}([a-zA-Z]+).([0-9]{2}|N/A)\\s\\1"; // changes to [0-9], now OK
regcomp(&re, pattern, REG_EXTENDED|REG_NOSUB);
ret = regexec(&re, target, (size_t) 0, NULL, 0);
if(ret == REG_NOMATCH) {
printf("\n%s\n does not match \n%s\n", target, pattern);
}
regfree(&re);
return 0;
}
I have read the POSIX regular expressions standard on Wikipedia, I haven't found any evidence for that whether the POSIX stand suports the \d metacharacter.
Or there is a subtle mistake in my code?