Regex \ w escaping in C ++ 11

This code returns nothing, I am avoiding the w character in the wrong way

http://liveworkspace.org/code/3bRWOJ $ 38

#include <iostream>
#include <regex>
using namespace std;



int main()
{
    const char *reg_esp = "\w";  // List of separator characters.

// this can be done using raw string literals:
// const char *reg_esp = R"([ ,.\t\n;:])";

std::regex rgx(reg_esp); // 'regex' is an instance of the template class
                         // 'basic_regex' with argument of type 'char'.
std::cmatch match; // 'cmatch' is an instance of the template class
                   // 'match_results' with argument of type 'const char *'.
const char *target = "Unseen University - Ankh-Morpork";

// Identifies all words of 'target' separated by characters of 'reg_esp'.
if (std::regex_search(target, match, rgx)) {
    // If words separated by specified characters are present.

    const size_t n = match.size();
    for (size_t a = 0; a < n; a++) {
        std::string str (match[a].first, match[a].second);
        std::cout << str << "\n";
    }
}

    return 0;
}

      

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2 answers


The regex must contain \w

two characters, \

and w

therefore your C ++ source code must contain "\\w"

, since you need to escape the backslash.



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As @DanielFrey said, for a normal string literal, you need to double the backslash. Since C ++ 11, a string literal: can be used instead R"(\w)"

. "R" disables special character handling, so the backslash is just a backslash. The parentheses indicate the beginning and end of a string literal and are not part of the text.



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