如何使用 C 正则表达式提取子匹配

Question

I have made GNU regex library work exactly as advertised in an extensive text processing alg that I wrote about 2 years ago, but unfortunately that platform is gone and I don't know whether its versions were older or newer than that referenced below.

Here is the code:

// GNU libc version: 2.28
// gcc (Ubuntu 7.5.0-3ubuntu1~18.04) 7.5.0

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

int main() {

    regex_t preg;
    char str[] = "dave";
    char regex[] = "\\(.\\)ave";

    // flag REG_EXTENDED with unescaped parens in the r.e. doesn't fix anything
    int ret, cflags = REG_ICASE;
    
    // the elements of unused pmatches used to be set to -1 by regexec, but no longer. a clue perhaps.

    regmatch_t pmatch[2] = {{-1,-1},{-1,-1}};

    ret = regcomp(&preg, regex, cflags);
    if (ret) {
        puts("regcomp fail");
        return ret;
    }
    else
        // preg.re_nsub contains the correct number of groups that regcomp recognized in the r.e. Tests succeeded for 0, 1, 2, and 3 groups.
        printf("regcomp ok; re_nsub=%zu\n", preg.re_nsub);

    ret = regexec(&preg, str, 1, pmatch, 0);

    if(ret)
        puts("no match");
    else {
        printf("match offsets are %d %d\n", pmatch[0].rm_so, pmatch[0].rm_eo);
        printf("match[0]=%*s<\n", pmatch[0].rm_eo, &str[pmatch[0].rm_so]);

        printf("submatch offsets are %d %d\n", pmatch[1].rm_so, pmatch[1].rm_eo);
        if(pmatch[1].rm_so != -1)
            printf("match[1]=%*s<\n", pmatch[1].rm_eo, &str[pmatch[1].rm_so]);
    }
    return 0;
}
/*
output:
regcomp ok; re_nsub=1
match offsets are 0 4
match[0]=dave<
submatch offsets are -1 -1
*/

Answer 1

The issue you do not get offsets for the first capturing group is that you pass 1 as the third, size_t __nmatch , argument to regexec .

The 1 value should be changed to 2 as there will be two groups whenever \(.\)ave regex matches: Group 0 will be holding the whole match and Group 1 will hold the first capturing group value.

So, you need to use

ret = regexec(&preg, str, 2, pmatch, 0);
//                       ^^^

Also, to print the Group 1 value you can use

if(pmatch[1].rm_so != -1) {
    printf("match[1]=%.*s<\n", pmatch[1].rm_eo, &str[pmatch[1].rm_so]);
}

See this C demo :

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

int main() {

    regex_t preg;
    char str[] = "dave";
    char regex[] = "\\(.\\)ave";

    // flag REG_EXTENDED with unescaped parens in the r.e. doesn't fix anything
    int ret, cflags = REG_ICASE;
    
    // the elements of unused pmatches used to be set to -1 by regexec, but no longer. a clue perhaps.

    regmatch_t pmatch[2] = {{-1,-1},{-1,-1}};

    ret = regcomp(&preg, regex, cflags);
    if (ret) {
        puts("regcomp fail");
        return ret;
    }
    else
        // preg.re_nsub contains the correct number of groups that regcomp recognized in the r.e. Tests succeeded for 0, 1, 2, and 3 groups.
        printf("regcomp ok; re_nsub=%zu\n", preg.re_nsub);

    ret = regexec(&preg, str, 2, pmatch, 0); // 1 changed to 2 as there is Group 0 (whole match) and Group 1 (for the first capturing group)

    if(ret)
        puts("no match");
    else {
        printf("match offsets are %d %d\n", pmatch[0].rm_so, pmatch[0].rm_eo);
        printf("match[0]=%*s<\n", pmatch[0].rm_eo, &str[pmatch[0].rm_so]);

        printf("submatch offsets are %d %d\n", pmatch[1].rm_so, pmatch[1].rm_eo);
        if(pmatch[1].rm_so != -1) {
            printf("match[1]=%.*s<\n", pmatch[1].rm_eo, &str[pmatch[1].rm_so]);
        }
    }
    return 0;
}
/*
regcomp ok; re_nsub=1
match offsets are 0 4
match[0]=dave<
submatch offsets are 0 1
match[1]=d<
*/