"Jorge Lodos Vigil" <lodos@segurmatica.cu> wrote in message news:ECBF993526E3BC47BD232BDC77D6933706A916F4E9@mercurio.segurmatica.cu...
Hi We are using xpressive with a grammar to match certain patterns. In some cases, we have the need to ignore white spaces. Using dynamic regexes, this can be achieved with the ignore_white_space constant. Is there a way to ignore white spaces using a grammar in xpressive other than modifying the grammar itself?
How about using a filter_iterator to skip spaces. Something like the following program: #include <iostream> #include <string> #include <boost/config.hpp> #include <boost/iterator/filter_iterator.hpp> #include <boost/range/iterator_range.hpp> #include <boost/xpressive/xpressive_static.hpp> struct not_space { inline bool operator()(char ch) const { return ' ' != ch; } }; typedef boost::filter_iterator<not_space, std::string::const_iterator> Iter_Skip; boost::iterator_range<Iter_Skip> make_range (std::string& s) { return boost::make_iterator_range( boost::make_filter_iterator<not_space>(s.begin(), s.end()), boost::make_filter_iterator<not_space>(s.end(), s.end()) ); } int main() { using namespace boost::xpressive; std::string s = "aa bb cc"; typedef basic_regex<Iter_Skip> regex_skip; regex_skip rx = as_xpr("aa") >> "bbcc"; match_results<Iter_Skip> what; if(!regex_match(make_range(s), what, rx)) std::cout << "not found\n"; else std::cout << "found\n"; return 0; }