const auto input = "Some people, when confronted with a problem, think \"I know, I'll use regular expressions.\""s;
smatch sm;

cout << input << endl;

// If input ends in a quotation that contains a word that begins with "reg" and another word begining with "ex" then capture the preceeding portion of input
if (regex_match(input, sm, regex("(.*)\".*\\breg.*\\bex.*\"\\s*$"))) {
    const auto capture = sm[1].str();
    
    cout << '\t' << capture << endl; // Outputs: "\tSome people, when confronted with a problem, think\n"
    
    // Search our capture for "a problem" or "# problems"
    if(regex_search(capture, sm, regex("(a|d+)\\s+problems?"))) {
        const auto count = sm[1] == "a"s ? 1 : stoi(sm[1]);
        
        cout << '\t' << count << (count > 1 ? " problems\n" : " problem\n"); // Outputs: "\t1 problem\n"
        cout << "Now they have " << count + 1 << " problems.\n"; // Ouputs: "Now they have 2 problems\n"
    }
}

Live Example

This code takes in various brace styles and converts them to One True Brace Style:

const auto input = "if (KnR)\n\tfoo();\nif (spaces) {\n    foo();\n}\nif (allman)\n{\n\tfoo();\n}\nif (horstmann)\n{\tfoo();\n}\nif (pico)\n{\tfoo(); }\nif (whitesmiths)\n\t{\n\tfoo();\n\t}\n"s;

cout << input << regex_replace(input, regex("(.+?)\\s*\\{?\\s*(.+?;)\\s*\\}?\\s*"), "$1 {\n\t$2\n}\n") << endl;

Live Example

A std::regex_token_iterator provides a tremendous tool for extracting elements of a Comma Separated Value file. Aside from the advantages of iteration, this iterator is also able to capture escaped commas where other methods struggle:

const auto input = "please split,this,csv, ,line,\\,\n"s;
const regex re{ "((?:[^\\\\,]|\\\\.)+)(?:,|$)" };
const vector<string> m_vecFields{ sregex_token_iterator(cbegin(input), cend(input), re, 1), sregex_token_iterator() };

cout << input << endl;

copy(cbegin(m_vecFields), cend(m_vecFields), ostream_iterator<string>(cout, "\n"));

Live Example

A notable gotcha with regex iterators is, that the regex argument must be an L-value. An R-value will not work.

When processing of captures has to be done iteratively a regex_iterator is a good choice. Dereferencing a regex_iterator returns a match_result. This is great for conditional captures or captures which have interdependence. Let's say that we want to tokenize some C++ code. Given:

enum TOKENS {
    NUMBER,
    ADDITION,
    SUBTRACTION,
    MULTIPLICATION,
    DIVISION,
    EQUALITY,
    OPEN_PARENTHESIS,
    CLOSE_PARENTHESIS
};

We can tokenize this string: const auto input = "42/2 + -8\t=\n(2 + 2) * 2 * 2 -3"s with a regex_iterator like this:

vector<TOKENS> tokens;
const regex re{ "\\s*(\\(?)\\s*(-?\\s*\\d+)\\s*(\\)?)\\s*(?:(\\+)|(-)|(\\*)|(/)|(=))" };

for_each(sregex_iterator(cbegin(input), cend(input), re), sregex_iterator(), [&](const auto& i) {
    if(i[1].length() > 0) {
        tokens.push_back(OPEN_PARENTHESIS);
    }
    
    tokens.push_back(i[2].str().front() == '-' ? NEGATIVE_NUMBER : NON_NEGATIVE_NUMBER);
    
    if(i[3].length() > 0) {
        tokens.push_back(CLOSE_PARENTHESIS);
    }        
    
    auto it = next(cbegin(i), 4);
    
    for(int result = ADDITION; it != cend(i); ++result, ++it) {
        if (it->length() > 0U) {
            tokens.push_back(static_cast<TOKENS>(result));
            break;
        }
    }
});

match_results<string::const_reverse_iterator> sm;

if(regex_search(crbegin(input), crend(input), sm, regex{ tokens.back() == SUBTRACTION ? "^\\s*\\d+\\s*-\\s*(-?)" : "^\\s*\\d+\\s*(-?)" })) {
    tokens.push_back(sm[1].length() == 0 ? NON_NEGATIVE_NUMBER : NEGATIVE_NUMBER);
}

Live Example

A notable gotcha with regex iterators is that the regex argument must be an L-value, an R-value will not work: Visual Studio regex_iterator Bug?

std::vector<std::string> split(const std::string &str, std::string regex)
{
    std::regex r{ regex };
    std::sregex_token_iterator start{ str.begin(), str.end(), r, -1 }, end;
    return std::vector<std::string>(start, end);
}

split("Some  string\t with whitespace ", "\\s+"); // "Some", "string", "with", "whitespace"

Let's say that we're given const string input as a phone number to be validated. We could start by requiring a numeric input with a zero or more quantifier: regex_match(input, regex("\\d*")) or a one or more quantifier: regex_match(input, regex("\\d+")) But both of those really fall short if input contains an invalid numeric string like: "123" Let's use a n or more quantifier to ensure that we're getting at least 7 digits:

regex_match(input, regex("\\d{7,}"))

This will guarantee that we will get at least a phone number of digits, but input could also contain a numeric string that's too long like: "123456789012". So lets go with a between n and m quantifier so the input is at least 7 digits but not more than 11:

regex_match(input, regex("\\d{7,11}"));

This gets us closer, but illegal numeric strings that are in the range of [7, 11] are still accepted, like: "123456789" So let's make the country code optional with a lazy quantifier:

regex_match(input, regex("\\d?\\d{7,10}"))

It's important to note that the lazy quantifier matches as few characters as possible, so the only way this character will be matched is if there are already 10 characters that have been matched by \d{7,10}. (To match the first character greedily we would have had to do: \d{0,1}.) The lazy quantifier can be appended to any other quantifier.

Now, how would we make the area code optional and only accept a country code if the area code was present?

regex_match(input, regex("(?:\\d{3,4})?\\d{7}"))

In this final regex, the \d{7} requires 7 digits. These 7 digits are optionally preceded by either 3 or 4 digits.

Note that we did not append the lazy quantifier: \d{3,4}?\d{7}, the \d{3,4}? would have matched either 3 or 4 characters, preferring 3. Instead we're making the non-capturing group match at most once, preferring not to match. Causing a mismatch if input didn't include the area code like: "1234567".

---

In conclusion of the quantifier topic, I'd like to mention the other appending quantifier that you can use, the possessive quantifier. Either the lazy quantifier or the possessive quantifier can be appended to any quantifier. The possessive quantifier's only function is to assist the regex engine by telling it, greedily take these characters and don't ever give them up even if it causes the regex to fail. This for example doesn't make much sense: regex_match(input, regex("\\d{3,4}+\\d{7})) Because an input like: "1234567890" wouldn't be matched as \d{3,4}+ will always match 4 characters even if matching 3 would have allowed the regex to succeed.
The possessive quantifier is best used when the quantified token limits the number of matchable characters. For example:

regex_match(input, regex("(?:.*\\d{3,4}+){3}"))

Can be used to match if input contained any of the following:

123 456 7890
123-456-7890
(123)456-7890
(123) 456 - 7890

But when this regex really shines is when input contains an illegal input:

12345 - 67890

Without the possessive quantifier the regex engine has to go back and test every combination of .* and either 3 or 4 characters to see if it can find a matchable combination. With the possessive quantifier the regex starts where the 2^nd possessive quantifier left off, the '0' character, and the regex engine tries to adjust the .* to allow \d{3,4} to match; when it can't the regex just fails, no back tracking is done to see if earlier .* adjustment could have allowed a match.

C++ provides only 4 anchors:

• ^ which asserts the start of the string
• $ which asserts the end of the string
• \b which asserts a \W character or the beginning or end of the string
• \B which asserts a \w character

Let's say for example we want to capture a number with it's sign:

auto input = "+1--12*123/+1234"s;
smatch sm;

if(regex_search(input, sm, regex{ "(?:^|\\b\\W)([+-]?\\d+)" })) {

    do {
        cout << sm[1] << endl;
        input = sm.suffix().str();
    } while(regex_search(input, sm, regex{ "(?:^\\W|\\b\\W)([+-]?\\d+)" }));
}

Live Example

An important note here is that the anchor does not consume any characters.