#include <regex.h> for problems with creating correct regexp

Question

I would like to grab everything before ] in aasd{123;'asd'aaaa]asd. So I tried to create regex but regex.h works weird.

Closest I get: ".*[^\\]]" which gives me: aasd{123;'asd'aaaa]

#include <regex.h>
#include <iostream>
#include <cstring>

using namespace std;

int main(){


    const char* phrase = "aasd{123;'asd'aaaa]asd";
    const char* expression = "........"; // <- what should be here? 

    size_t group_size = 5;
    regmatch_t *pmatch = new regmatch_t[group_size];
    memset(pmatch, 0, sizeof(regmatch_t) * group_size);

    regex_t regex;

    int reti = regcomp(&regex, expression, REG_EXTENDED);

    if (reti){
        cout << "Error1" << reti <<endl;
        return false;
    }

    reti = regexec(&regex, phrase, group_size, pmatch, 0);
    if (reti){
        cout << "Error2 " << reti <<endl;
        return false;
    }

    int len = 0, i = 0;
    char result[10000];

    for (i = 0; pmatch[i].rm_so != -1; i++){
        len = pmatch[i].rm_eo - pmatch[i].rm_so;
        memcpy(&result, phrase + pmatch[i].rm_so, len);
        cout << "num " << i << " " << result << endl;
   }
}

Jeffery Thomas · Accepted Answer · 2014-02-05 11:38:02Z

1

^[^\\]]* is what you are looking for.

In english: from the start of the line (^) match everything which is not a "]" ([^\\]]*).

UPDATE

There must be an odd quirk to this library, ^[^]]* passed the test.

The "^[^\\]]*" literal get complied into ^[^\]]*. This should have been an escaped ] in a character class [], but this library turned it into a character class containing not \ followed by any number of ]s.

^ [^\] ]* instead of ^ [^\]]*.

Fortunately, there is an alternate form which allows for []] to match ] if it's the only character in the character class. This extended to [^]] to match a non ].

In short… try ^[^]]*.

edited Feb 5, 2014 at 11:38

answered Feb 5, 2014 at 3:17

Jeffery Thomas

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2 Comments

zie1ony Over a year ago

Just paste your regexp in my code as const char* expression = "^[^\\]]*";

Jeffery Thomas Over a year ago

@zie1ony heh, odd quirk of this library. I updated my answer.

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