Boost Spirit综合属性混淆

Question

我试图解析输入，它有一个加号或减号字符，后跟一个X或Y字符，后跟一个无符号整数。

(char_('+') | char_('-')) >> char_("xyXY") >> uint_ 

根据我的文档的读数，此合成的属性将是tuple<vector<char>,unsigned int>因为替代解析器(char | char)将char型时，char >> char("xyXY")将vector<char>，并且vector<char> >> uint_将是类型的元组，所以tuple<vector<char>,unsigned int>。这编译失败

qi\detail\assign_to.hpp(152) : error C2440: 'static_cast' : cannot convert from 'const char' to 'boost::tuples::tuple<T0,T1>' 

代码：

#include <iostream> 
#include <string> 
#include <vector> 

#include <boost/fusion/include/tuple.hpp> 
#include <boost/spirit/include/qi.hpp> 
#include <boost/tuple/tuple.hpp> 

using namespace boost::spirit::qi; 

int main() 
{ 
    std::string input("-Y 512"); 
    typedef std::string::const_iterator Iterator; 
    Iterator first = input.begin(); 
    Iterator last = input.end(); 
    boost::tuple<std::vector<char>,unsigned int> output; 
    bool result = phrase_parse(first,last,(char_('+') | char_('-')) >> char_("xyXY") >> uint_,ascii::space,output); 
    if(result && first == last) 
     std::cout << "sign=" << boost::get<0>(output)[0] << ", xy=" << boost::get<0>(output)[1] << ", size=" << boost::get<1>(output) << '\n'; 
    else 
     std::cerr << "Parse error\n"; 
} 

我又试图tuple<char,char,unsigned int>的属性类型：

#include <iostream> 
#include <string> 
#include <vector> 

#include <boost/fusion/include/tuple.hpp> 
#include <boost/spirit/include/qi.hpp> 
#include <boost/tuple/tuple.hpp> 

using namespace boost::spirit::qi; 

int main() 
{ 
    std::string input("-Y 512"); 
    typedef std::string::const_iterator Iterator; 
    Iterator first = input.begin(); 
    Iterator last = input.end(); 
    boost::tuple<char,char,unsigned int> output; 
    bool result = phrase_parse(first,last,(char_('+') | char_('-')) >> char_("xyXY") >> uint_,ascii::space,output); 
    if(result && first == last) 
     std::cout << "sign=" << boost::get<0>(output) << ", xy=" << boost::get<1>(output) << ", size=" << boost::get<2>(output) << '\n'; 
    else 
     std::cerr << "Parse error\n"; 
} 

这编译，但输出不正确。输入的所述第一令牌正确分析，但随后的令牌不是：

sign=-, xy= , size=0 

我还试图as_string[]：

#include <iostream> 
#include <string> 
#include <vector> 

#include <boost/fusion/include/tuple.hpp> 
#include <boost/spirit/include/qi.hpp> 
#include <boost/tuple/tuple.hpp> 

using namespace boost::spirit::qi; 

int main() 
{ 
    std::string input("-Y 512"); 
    typedef std::string::const_iterator Iterator; 
    Iterator first = input.begin(); 
    Iterator last = input.end(); 
    boost::tuple<std::string,unsigned int> output; 
    bool result = phrase_parse(first,last,as_string[(char_('+') | char_('-')) >> char_("xyXY")] >> uint_,ascii::space,output); 
    if(result && first == last) 
     std::cout << "sign=" << boost::get<0>(output)[0] << ", xy=" << boost::get<0>(output)[1] << ", size=" << boost::get<1>(output) << '\n'; 
    else 
     std::cerr << "Parse error\n"; 
} 

这种改进的东西作为X/Y令牌得到解析，但是不是第三个整数标记：

sign=-, xy=Y, size=0 

请告诉我我哪里出错了。

我使用的是精神版本2.5.2（与Boost 1.58.0）和Microsoft Visual Studio 2008

Answer 1

Spirit库文档推荐使用Fusion tuple。我想我看到了某处（现在找不到它），Boost tuple可能与Spirit库不完全兼容。

这里是你的固定例如：

#include <iostream> 
#include <string> 
#include <vector> 

#include <boost/fusion/include/tuple.hpp> 
#include <boost/spirit/include/qi.hpp> 
#include <boost/fusion/sequence.hpp> 

namespace qi = boost::spirit::qi; 

int main() 
{ 
    std::string input("-Y 512"); 
    typedef std::string::const_iterator Iterator; 
    Iterator first = input.begin(); 
    Iterator last = input.end(); 

    boost::fusion::tuple<char, char, unsigned int> output; 
    bool result = qi::phrase_parse(first, last, (qi::char_('+') | qi::char_('-')) >> qi::char_("xyXY") >> qi::uint_, qi::ascii::space, output); 
    if (result && first == last) 
     std::cout << "sign=" << boost::fusion::get<0>(output) << ", xy=" << boost::fusion::get<1>(output) << ", size=" << boost::fusion::get<2>(output) << '\n'; 
    else 
     std::cerr << "Parse error\n"; 

    return 0; 
} 

输出： sign=-, xy=Y, size=512

更新：其实我发现here，它可能使用boost::tuple，但不同的页眉需要包括：#include <boost/fusion/include/boost_tuple.hpp>。