Boost Spirit綜合屬性混淆

Question

我試圖解析輸入，它有一個加號或減號字符，後跟一個X或Y字符，後跟一個無符號整數。

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

根據我的文檔的讀數，此合成的屬性將是tuple<vector<char>,unsigned int>因爲替代解析器(char | char)將char型時，char >> char("xyXY")將vector<char>，並且vector<char> >> uint_將是類型的元組，所以tuple<vector<char>,unsigned int>。這編譯失敗

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

代碼：

#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"; 
} 

我又試圖tuple<char,char,unsigned int>的屬性類型：

#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"; 
} 

這編譯，但輸出不正確。輸入的所述第一令牌正確分析，但隨後的令牌不是：

sign=-, xy= , size=0 

我還試圖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"; 
} 

這種改進的東西作爲X/Y令牌得到解析，但是不是第三個整數標記：

sign=-, xy=Y, size=0 

請告訴我我哪裏出錯了。

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

Answer 1

Spirit庫文檔推薦使用Fusion tuple。我想我看到了某處（現在找不到它），Boost tuple可能與Spirit庫不完全兼容。

這裏是你的固定例如：

#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; 
} 

輸出： sign=-, xy=Y, size=512

更新：其實我發現here，它可能使用boost::tuple，但不同的頁眉需要包括：#include <boost/fusion/include/boost_tuple.hpp>。