c++ - 子解析器属性

Question

我在编写语法时遇到问题。假设我有一个继承自 Base 的类 Derived。 GrammarDerived 具有 Derived 综合属性，而 GrammarBase 具有 Base 综合属性。如何在 GrammarDerived 解析规则中使用 GrammarBase？我觉得这应该是可能的，因为我可以将 Base & 绑定(bind)到 Derived & 但似乎没有任何效果。

换句话说，如何让 grammarBase 通过下面的引用与 _val 交互？

template<typename Iterator>
struct GrammarDerived : public grammar <Iterator, Derived()> {
    GrammarDerived() : GrammarDerived::base_type(start) {
        start = rule1[bind(someFunc, _val)] >> grammarBase;
        rule1 = /* ... */;
    }
    rule<Iterator, Derived()> start;
    rule<Iterator, Derived()> rule1;
    GrammarBase grammarBase;
};

Answer 1

在一个更简单的设置中，这显示了这里主要是类型推导的限制:

Derived parse_result;
bool ok = qi::phrase_parse(f, l, base_, qi::space, data);

当解析器公开一个 Base 时将不起作用，但是您可以使用模板实例的“类型提示”修复它^[1]:

bool ok = qi::phrase_parse(f, l, base_, qi::space, static_cast<Base&>(data));

完整演示 Live On Coliru

#include <algorithm>
#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>

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

struct Base {
    int x;
    double y;
};

BOOST_FUSION_ADAPT_STRUCT(Base, (int,x)(double,y))

struct Derived : Base { };

int main()
{
    typedef std::string::const_iterator It;
    qi::rule<It, Base(), qi::space_type> base_ = qi::int_ >> qi::double_;

    std::string const input = "1 3.14";
    auto f(input.begin()), l(input.end());

    Derived parse_result;
    bool ok = qi::phrase_parse(f, l, base_, qi::space, static_cast<Base&>(parse_result));
    if (ok)
    {
        std::cout << "Parsed: " << parse_result.x << " " << parse_result.y << "\n";
    } else
    {
        std::cout << "Parse failed\n";
    }

    if (f != l)
    {
        std::cout << "Input remaining: '" << std::string(f,l) << "'\n";
    }
}

或者

您可以通过将对可公开属性的引用显式传递给基本解析器/规则来避免混淆:

template <typename It, typename Skipper = qi::space_type>
struct derived_grammar : qi::grammar<It, Derived(), Skipper>
{
    derived_grammar() : derived_grammar::base_type(start) {
        base_ = qi::int_ >> qi::double_;
        glue_ = base_ [ qi::_r1 = qi::_1 ];
        start = "derived:" >> glue_(qi::_val); // passing the exposed attribute for the `Base&` reference
    }
  private:
    qi::rule<It, Derived(),   Skipper> start;
    qi::rule<It, void(Base&), Skipper> glue_;
    qi::rule<It, Base(),      Skipper> base_; // could be a grammar instead of a rule
};

如果你真的坚持，你可以不用glue_/base_使用 qi::attr_cast<Base, Base> 分离(但为了易读性我不会这样做)。

完整代码再次引用 Live On Coliru

#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <algorithm>
#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>

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

struct Base {
    int x;
    double y;
};

BOOST_FUSION_ADAPT_STRUCT(Base, (int,x)(double,y))

struct Derived : Base { };

template <typename It, typename Skipper = qi::space_type>
struct derived_grammar : qi::grammar<It, Derived(), Skipper>
{
    derived_grammar() : derived_grammar::base_type(start) {
        base_ = qi::int_ >> qi::double_;
        glue_ = base_ [ qi::_r1 = qi::_1 ];
        start = "derived:" >> glue_(qi::_val); // passing the exposed attribute for the `Base&` reference
    }
  private:
    qi::rule<It, Derived(),   Skipper> start;
    qi::rule<It, void(Base&), Skipper> glue_;
    qi::rule<It, Base(),      Skipper> base_; // could be a grammar instead of a rule
};

int main()
{
    typedef std::string::const_iterator It;
    derived_grammar<It> g;

    std::string const input = "derived:1 3.14";
    auto f(input.begin()), l(input.end());

    Derived parse_result;
    bool ok = qi::phrase_parse(f, l, g, qi::space, parse_result);
    if (ok)
    {
        std::cout << "Parsed: " << parse_result.x << " " << parse_result.y << "\n";
    } else
    {
        std::cout << "Parse failed\n";
    }

    if (f != l)
    {
        std::cout << "Input remaining: '" << std::string(f,l) << "'\n";
    }
}

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^[1] 引用函数模板的实例化 qi::phrase_parse这里