C++ Boost: Split String - c++

Question

How can I split a string with Boost with a regex AND have the delimiter included in the result list?
for example, if I have the string "1d2" and my regex is "[a-z]" I want the results in a vector with (1, d, 2)
I have:
std::string expression = "1d2";
boost::regex re("[a-z]");
boost::sregex_token_iterator i (expression.begin (),
expression.end (),
re);
boost::sregex_token_iterator j;
std::vector <std::string> splitResults;
std::copy (i, j, std::back_inserter (splitResults));
Thanks

Answer 1

I think you cannot directly extract the delimiters using boost::regex. You can, however, extract the position where the regex is found in your string:
std::string expression = "1a234bc";
boost::regex re("[a-z]");
boost::sregex_iterator i(
expression.begin (),
expression.end (),
re);
boost::sregex_iterator j;
for(; i!=j; ++i) {
std::cout << (*i).position() << " : " << (*i) << std::endl;
}
This example would show:
1 : a
5 : b
6 : c
Using this information, you can extract the delimitiers from your original string:
std::string expression = "1a234bc43";
boost::regex re("[a-z]");
boost::sregex_iterator i(
expression.begin (),
expression.end (),
re);
boost::sregex_iterator j;
size_t pos=0;
for(; i!=j;++i) {
std::string pre_delimiter = expression.substr(pos, (*i).position()-pos);
std::cout << pre_delimiter << std::endl;
std::cout << (*i) << std::endl;
pos = (*i).position() + (*i).size();
}
std::string last_delimiter = expression.substr(pos);
std::cout << last_delimiter << std::endl;
This example would show:
1
a
234
b
c
43
There is an empty string betwen b and c because there is no delimiter.