Re: find function of String

Thank you for all your responses. I guess I am not very clear in my
question. Actually it is nothing complicated. All text except [ or ]
needs to be extracted. there is no complex stuff of subsets etc...

Anywayz I found a simple way using boost::regex to match only the
characters followed by numbers

int main(int argc, char* argv[])
{
    try
    {
        cout<< "[START REGEXP CHECKING]"<< endl;

        // String to be validated
        string myString = "[SBD 65 [INI 25 [PUP 0 ][RPN 25 ]][ANA 1 ][BUIPNR
0 ][BUIPAX 1 ]";

        // Regular Expression to be used.
        string myRegexp = "[a-zA-Z]+\\s\\d+";
        boost::regex aMyRegExp(myRegexp);

        // Needed to extract some part from myString, considering the sub-
expressions grouped by "()"
        boost::cmatch matches;

        cout<< endl<< "String: "<< myString;
        cout<< endl<< "Regular Expression: "<< myRegexp<< endl;

        // Checking if myString matches the Regular Expression.
        // Use boost::regex_match(myString.c_str(), aMyRegExp) if the
extraction is not needed
        std::string::const_iterator start, end;
        start = myString.begin();
        end = myString.end();
        boost::match_results<std::string::const_iterator> what;
        boost::match_flag_type flags = boost::match_default;
        while (regex_search(start, end, what, aMyRegExp, flags)){
            cout<< "Size: "<< what.size()<< endl;
            cout<< "Result: "<< what<< endl;

            // update start position
            start = what[0].second;
        }

        cout<< endl<< endl<< "[END REGEXP CHECKING]"<< endl;
        return 0;
    }
    catch (...)
    {
        cout<< "EXCEPTION!!!!"<< endl;
        return 1;
    }

}

This prints out only text in the format of XYZ 34 etc