使用簡單的Boost :: Spirit語法？

Question

我無法獲得語法工作，所以我簡化了它，直到它只解析一個整數。仍然無法讓它工作。這是下面的語法：

template<typename Iterator> 
struct rangeGrammar : qi::grammar<Iterator, int()> 
{ 
    rangeGrammar() : 
    rangeGrammar::base_type(number) 
    { 
     using qi::int_; 
     using qi::_1; 
     using qi::_val; 

     number = int_[_val = _1]; 
    } 
    qi::rule<Iterator, int()> number; 
}; 

它應該只是解析一個整數（我知道我可以只告訴解析功能使用int_的語法，但我wan't知道這個是錯的例）。

我的解析函數爲：

/* n is a std::string provided by the user */ 
rangeGrammar<std::string::const_iterator> grammar; 
int num = 0; 
qi::phrase_parse(n.start(), n.end(), grammar, num); 
std::cout << "Number: " << num << std::endl; 

我得到以下編譯器錯誤：

/boost/spirit/home/qi/reference.hpp: In member function ‘bool boost::spirit::qi::reference::parse(Iterator&, const Iterator&, Context&, const Skipper&, Attribute&) const [with Iterator = __gnu_cxx::__normal_iterator >, Context = boost::spirit::context, boost::spirit::locals<> >, Skipper = boost::spirit::unused_type, Attribute = int, Subject = const boost::spirit::qi::rule<__gnu_cxx::__normal_iterator >, int(), boost::spirit::unused_type, boost::spirit::unused_type, boost::spirit::unused_type>]’: /boost/spirit/home/qi/parse.hpp:89:82: instantiated from ‘bool boost::spirit::qi::parse(Iterator&, Iterator, const Expr&, Attr&) [with Iterator = __gnu_cxx::__normal_iterator >, Expr = rangeGrammar<__gnu_cxx::__normal_iterator > >, Attr = int]’ ../parameter_parser.h:95:46: instantiated from here boost/spirit/home/qi/reference.hpp:43:71: error: no matching function for call to ‘boost::spirit::qi::rule<__gnu_cxx::__normal_iterator >, int(), boost::spirit::unused_type, boost::spirit::unused_type, boost::spirit::unused_type>::parse(__gnu_cxx::__normal_iterator >&, const __gnu_cxx::__normal_iterator >&, boost::spirit::context, boost::spirit::locals<> >&, const boost::spirit::unused_type&, int&) const’ cc1plus: warnings being treated as errors /boost/spirit/home/qi/reference.hpp:44:9: error: control reaches end of non-void function * exit status 1 *

想不通的問題是什麼。任何幫助將不勝感激。

Answer 1

答：有錯誤的語法 什麼 小，但你使用qi::phrase_parse這需要一個隊長。使用qi::parse並且問題消失。

注1：使用[_val=_1]在那裏是完全多餘的;規則無語義屬性享受自動屬性傳播。

注2：您可能需要使用氣::比賽做分析是這樣的：

#include <boost/spirit/include/qi_match.hpp> 

const std::string input = "1234"; 
std::istringstream iss(input); 
iss >> qi::match(qi::int_ [ std::cout << qi::_1 ]); 

最後爲了您的利益，這裏有一個骷髏「doParse」與運行測試顯示功能好齊實踐一些要素：

#include <boost/spirit/include/qi.hpp> 
#include <boost/spirit/include/phoenix.hpp> 

namespace qi = boost::spirit::qi; 
namespace phx = boost::phoenix; 

template<typename Iterator> 
struct rangeGrammar : qi::grammar<Iterator, int()> 
{ 
    rangeGrammar() : rangeGrammar::base_type(number) 
    { 
     number = qi::int_; 
    } 
    qi::rule<Iterator, int()> number; 
}; 

bool doParse(const std::string& input) 
{ 
    typedef std::string::const_iterator It; 
    It f(begin(input)), l(end(input)); 

    try 
    { 
     rangeGrammar<It> p; 
     int data; 

     bool ok = qi::parse(f,l,p,data); 
     if (ok) 
     { 
      std::cout << "parse success\n"; 
      std::cout << "data: " << data << "\n"; 
     } 
     else  std::cerr << "parse failed: '" << std::string(f,l) << "'\n"; 

     if (f!=l) std::cerr << "trailing unparsed: '" << std::string(f,l) << "'\n"; 
     return ok; 
    } catch(const qi::expectation_failure<It>& e) 
    { 
     std::string frag(e.first, e.last); 
     std::cerr << e.what() << "'" << frag << "'\n"; 
    } 

    return false; 
} 

int main() 
{ 
    bool ok = doParse("1234"); 
    return ok? 0 : 255; 
}