Thread: regex.h - extracting matches

Strangely I haven't used regexp's much in C but I have kept this on file in case I have to do it again, if you want a working example:

Code:

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

char *regexp (char *string, char *patrn, int *begin, int *end) {     
        int i, w=0, len;                  
        char *word = NULL;
        regex_t rgT;
        regmatch_t match;
        regcomp(&rgT,patrn,REG_EXTENDED);
        if ((regexec(&rgT,string,1,&match,0)) == 0) {
                *begin = (int)match.rm_so;
                *end = (int)match.rm_eo;
                len = *end-*begin;
                word=malloc(len+1);
                for (i=*begin; i<*end; i++) {
                        word[w] = string[i];
                        w++; }
                word[w]=0;
        }
        regfree(&rgT);
        return word;
}


int main() {
	int b,e;
	char *match=regexp("this and7 that","[a-z]+[0-9]",&b,&e);
	printf("->%s<-\n(b=%d e=%d)\n",match,b,e);
	return 0;
}

beware that regexp.h uses POSIX style notation (no \d or \w, etc). Supposedly, there is stuff like [:digit:] but if I substitute this

Code:

char *match=regexp("this and7 that","[:alpha:]+[:digit:]",&b,&e);

on my system, instead of the predicatable:

->and7<-
(b=5 e=9)

I get the bizarre and inexplicable:

->hi<-
(b=1 e=3)

which does not look like a number of letters and a digit to me!

Originally Posted by cas

You want [[:digit:]].

Wow. I honestly don't
I mean that is only one character less than [0123456789]! Why not make it:

Code:

[[[[any single numeral, 1, 2, 3, 4, 5, 6, 7, 8, 9 and also including ZERO]]]]

But thanks.