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我编写了一些规则来将浮点数解析为两个 std::vector 的浮点数，它们又存储在一个结构中：

数据输入：

#
# object name01
#

v  -1.5701 33.8087 0.3592
v  -24.0119 0.0050 21.7439
# a comment

vn 0.0000 0.5346 0.8451
vn 0.8331 0.5531 -0.0000
# another comment

结构：

struct ObjParseData
{
    ObjParseData() : verts(), norms() {}

    std::vector<float> verts;
    std::vector<float> norms;
};

以及相关的解析代码：

struct objGram : qi::grammar<std::string::const_iterator, ObjParseData(), iso8859::space_type>
    {
        objGram() : objGram::base_type(start)
        {
            vertex  = 'v' >> qi::double_ >> qi::double_ >> qi::double_;
            normal  = "vn" >> qi::double_ >> qi::double_ >> qi::double_;
            comment = '#' >> qi::skip(qi::blank)[ *(qi::print) ];
            vertexList = *(vertex | comment);
            normalList = *(normal | comment);
            start = vertexList >> normalList;
        }

        qi::rule<std::string::const_iterator, ObjParseData(), iso8859::space_type> start;
        qi::rule<std::string::const_iterator, std::vector<float>(), iso8859::space_type> vertexList;
        qi::rule<std::string::const_iterator, std::vector<float>(), iso8859::space_type> normalList;
        qi::rule<std::string::const_iterator, std::vector<float>(), iso8859::space_type> vertex;
        qi::rule<std::string::const_iterator, std::vector<float>(), iso8859::space_type> normal;
        qi::rule<std::string::const_iterator, iso8859::space_type> comment;
    } objGrammar;


    ObjParseData resultData;

    std::string::const_iterator f = data.cbegin();
    bool res = qi::phrase_parse( f, data.cend(), objGrammar, iso8859::space, resultData );

这有效。它将所有以“v”开头的浮点数解析为结构的顶点向量，并将所有以“vn”开头的浮点数解析为范数。这很棒，但我真的不知道为什么会这样。

现在，如果我正确理解这一点，如下定义的规则会将其所有结果放入浮点数的 std::vector 中。

qi::rule<std::string::const_iterator, std::vector<float>(), iso8859::space_type> vertex;

因此，查看上面显示的解析代码并知道像 vertex 这样的规则会解析为 float 的 std::vector，显然像 vertexList （如上所示）这样的规则将来自 vertex 的结果连接到一个 std::vector 的 float ? 所以看到这种行为，你会认为你可以把这两个规则（顶点和顶点列表）写成一个，但不幸的是这不起作用：

vertex  = *('v' >> qi::double_ >> qi::double_ >> qi::double_) | comment;
normal  = *("vn" >> qi::double_ >> qi::double_ >> qi::double_) | comment;
comment = '#' >> qi::skip(qi::blank)[ *(qi::print) ];
start = vertex >> normal;

代码确实编译并且 qi::phrase_parse 确实返回了一个成功的解析，但是结构中的 std::vector 不再填充了。我在这里缺少什么？

4

1 回答 1

3

您放错了分组括号：扩展

    vertexList = *(vertex | comment);
    normalList = *(normal | comment);

通过消除子规则导致

    vertex     = *(('v'  >> qi::double_ >> qi::double_ >> qi::double_) | comment);
    normal     = *(("vn" >> qi::double_ >> qi::double_ >> qi::double_) | comment);

或者，我更喜欢：

完整的工作示例（请下次制作您的代码示例 SSCCE？https ://meta.stackexchange.com/questions/22754/sscce-how-to-provide-examples-for-programming-questions ）：

#include <iterator>
#include <fstream>
#include <boost/fusion/adapted.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/karma.hpp>
#include <boost/spirit/include/phoenix.hpp>

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

struct ObjParseData
{
    ObjParseData() : verts(), norms() {}

    std::vector<float> verts;
    std::vector<float> norms;
};

BOOST_FUSION_ADAPT_STRUCT(ObjParseData, (std::vector<float>, verts)(std::vector<float>, norms))



template <typename It, typename Skipper = qi::space_type>
    struct parser : qi::grammar<It, ObjParseData(), Skipper>
{
    parser() : parser::base_type(start)
    {
        using namespace qi;


        vertex     = 'v'  >> qi::double_ >> qi::double_ >> qi::double_;
        normal     = "vn" >> qi::double_ >> qi::double_ >> qi::double_;
        comment    = '#' >> qi::skip(qi::blank)[ *(qi::print) ];
#if 0
        vertexList = *(vertex | comment);
        normalList = *(normal | comment);
        start      = vertexList >> normalList;
#else
        vertex     = *(comment | ('v'  >> qi::double_ >> qi::double_ >> qi::double_));
        normal     = *(comment | ("vn" >> qi::double_ >> qi::double_ >> qi::double_));
        start      = vertex >> normal;                                              
#endif

        BOOST_SPIRIT_DEBUG_NODE(start);
    }

  private:
    qi::rule<std::string::const_iterator, ObjParseData(), qi::space_type> start;
    qi::rule<std::string::const_iterator, std::vector<float>(), qi::space_type> vertexList;
    qi::rule<std::string::const_iterator, std::vector<float>(), qi::space_type> normalList;
    qi::rule<std::string::const_iterator, std::vector<float>(), qi::space_type> vertex;
    qi::rule<std::string::const_iterator, std::vector<float>(), qi::space_type> normal;
    qi::rule<std::string::const_iterator, qi::space_type> comment;
};

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

    parser<It, qi::space_type> p;
    ObjParseData data;

    try
    {
        bool ok = qi::phrase_parse(f,l,p,qi::space,data);
        if (ok)   
        {
            std::cout << "parse success\n";
            std::cout << "data: " << karma::format_delimited(
                    "v: " << karma::auto_ << karma::eol <<
                    "n: " << karma::auto_ << karma::eol, ' ', data);
        }
        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()
{
    std::ifstream ifs("input.txt", std::ios::binary);
    ifs.unsetf(std::ios::skipws);
    std::istreambuf_iterator<char> f(ifs), l;

    bool ok = doParse({ f, l });
}

输出：

parse success
data: v:  -1.57 33.809 0.359 -24.012 0.005 21.744 
 n:  0.0 0.535 0.845 0.833 0.553 0.0
于 2013-07-11T17:04:49.793 回答