[英]boost spirit debug rule with locals
當my_rule具有一些自定義類型的局部變量時,我無法在調試模式下編譯代碼(BOOST_SPIRIT_DEBUG_NODE(my_rule)代碼)。
qi::locals<std::string>
第一個版本可以 qi::locals<std::string,int>
第二版仍然可以 qi::locals<std::string,std::vector<int> >
當前版本無法編譯。 錯誤: operator<<
不匹配(操作數類型為std::basic_ostream<char>
和const std::vector<int>
)
我聲明流operator<<
:
std::ostream& > operator<< (std::ostream& os, std::vector<int> const& art)
但是它仍然無法編譯。
我使用升壓1_64_0。 這是最小的完整代碼:
#define BOOST_SPIRIT_DEBUG
#if !defined(BOOST_SPIRIT_DEBUG_OUT)
#define BOOST_SPIRIT_DEBUG_OUT std::cerr
#endif
#include <tuple>
#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/include/phoenix_fusion.hpp>
#include <boost/spirit/include/phoenix_stl.hpp>
#include <boost/spirit/include/phoenix_object.hpp>
#include <iostream>
#include <string>
#include <vector>
// To solve the pb of declaration of grammar with locals
#include <typeinfo>
std::ostream&
operator<< (std::ostream& os, std::vector<int> const& art)
{
os << "[";
for( auto it = art.begin(); it != art.end() ; it++ ) {
os << *it << ",";
}
os << "]";
return os;
}
namespace client
{
namespace phoenix = boost::phoenix;
namespace qi = boost::spirit::qi;
namespace ascii = boost::spirit::ascii;
using phoenix::val;
using namespace qi::labels;
using qi::_val;
using qi::_1;
// Our number list parser
template <typename Iterator>
struct mini_wkart_grammar
// first version: : qi::grammar<Iterator, int(), qi::locals<std::string>, ascii::space_type>
// second version: : qi::grammar<Iterator, int(), qi::locals<std::string,int>, ascii::space_type>
: qi::grammar<Iterator, std::vector<int>(), qi::locals<std::string,std::vector<int> >, ascii::space_type>
{
mini_wkart_grammar() : mini_wkart_grammar::base_type(start,"numbers")
{
using phoenix::push_back;
// first version: start= (qi::int_ >> qi::char_(',') >> qi::int_)[_val=_1+_3];
// second version: start= (qi::int_[_b=_1] >> qi::char_(',') >> qi::int_[_b+=_1])[_val=_b];
start= (qi::int_[push_back(_b,_1)] >> qi::char_(',') >> qi::int_[push_back(_b,_1)])[_val=_b];
BOOST_SPIRIT_DEBUG_NODE(start);
}
// first version OK: qi::rule<Iterator, int(), qi::locals<std::string>, ascii::space_type> start;
// second version OK: qi::rule<Iterator, int(), qi::locals<std::string,int>, ascii::space_type> start;
qi::rule<Iterator, std::vector<int>(), qi::locals<std::string,std::vector<int> >, ascii::space_type> start;
};
}
////////////////////////////////////////////////////////////////////////////
// Main program
////////////////////////////////////////////////////////////////////////////
int
main()
{
std::cout << "/////////////////////////////////////////////////////////\n\n";
std::cout << "\t\tA comma separated list parser for Spirit...\n\n";
std::cout << "/////////////////////////////////////////////////////////\n\n";
std::cout << "Give me a comma separated list of numbers.\n";
std::cout << "Type [q or Q] to quit\n\n";
// std::string result;
// first ans second version: int result;
std::vector<int> result;
std::string str;
using boost::spirit::ascii::space;
client::mini_wkart_grammar<std::string::const_iterator> wkart_grammar;
while (getline(std::cin, str))
{
if (str.empty() || str[0] == 'q' || str[0] == 'Q')
break;
std::string::const_iterator iter = str.begin();
std::string::const_iterator end = str.end();
// if (client::parse_numbers(str.begin(), str.end()))
if (boost::spirit::qi::phrase_parse(iter, end, wkart_grammar, space, result))
{
std::cout << "-------------------------\n";
std::cout << "Parsing succeeded\n";
std::cout << result << " Parses OK: " << std::endl;
}
else
{
std::cout << "-------------------------\n";
std::cout << "Parsing failed\n";
std::cout << "-------------------------\n";
}
}
std::cout << "Bye... :-) \n\n";
return 0;
}
我認為操作員聲明中缺少某些內容?
謝謝你的幫助..
您到底想達到什么目的? 整個語法可以是start = qi::int_ % ',';
並且仍然具有完全相同的效果。 參見Boost Spirit:“語義行為是邪惡的”?
可悲的是,您需要使該operator<<
啟用ADL。 ( http://en.cppreference.com/w/cpp/language/adl )
由於元素類型是原始類型,因此沒有關聯的名稱空間。 因此,將嘗試使用的唯一命名空間是namespace ::std
,它聲明了std::vector<>
。
namespace std {
std::ostream &operator<<(std::ostream &os, vector<int> const &art) {
os << "[";
for (auto it = art.begin(); it != art.end(); it++) {
os << *it << ",";
}
os << "]";
return os;
}
}
這可能會產生不希望有的副作用,您可能需要使用hack來強制解決該問題:
namespace ADL_Hack {
template <typename T>
struct allocator : std::allocator<T> { };
}
template <typename T>
using Vector = std::vector<T, ADL_Hack::allocator<T> >;
namespace ADL_Hack {
template <typename... Ts>
std::ostream &operator<<(std::ostream &os, std::vector<Ts...> const &art) {
os << "[";
for (auto it = art.begin(); it != art.end(); it++) {
os << *it << ",";
}
os << "]";
return os;
}
}
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