Question

问题在于我制作了一个对某项任务有用的语法，但现在却是任务已经改变，我需要定义新规则。

但是我不想修改我已经拥有的语法而不是我的语法喜欢创建一个新的语法，使用我没有代码的现有语法重复，所以我只需要定义我需要的新规则。我试了一下像这样，但不起作用：

struct New_grammar : Old_grammar<Iterator, Skipper>    
{
    New_grammar() : New_grammar::base_type(Command_list)
    {
        Command_list %= qi::eps >> + Commands;
        Comandos %= oneoldCommand | NewCommand;
        NewCommand = ("NewCommand" >> stmt)[qi::_val = phoenix::new_<NewCom>(qi::_1)];
    }
    // this is a new rule I need:
    qi::rule<Iterator, Commands*(), qi::locals<std::string>, Skipper> NewCommand; 
};

基本上Old_grammar是我已经拥有的语法，我只想添加我在New_grammar中需要的新规则，也可以使用规则和我已经在Old_gramar中的语法。

Answer 1

我不会通过继承来使问题复杂化。组合通常绰绰有余，并且不会混淆qi解析器接口。

我已经制作了一个关于如何完成版本化语法的小草图。假设旧语法：

template <typename It, typename Skipper>
struct OldGrammar : qi::grammar<It, Skipper, std::string()>
{
    OldGrammar() : OldGrammar::base_type(mainrule)
    {
        using namespace qi;
        rule1 = int_(1); // expect version 1
        rule2 = *char_;  // hopefully some interesting grammar
        mainrule = omit [ "version" > rule1 ] >> rule2;
    }
  private:
    qi::rule<It, Skipper, std::string()> mainrule;
    qi::rule<It, Skipper, int()>         rule1;
    qi::rule<It, Skipper, std::string()> rule2;
};

正如您所看到的，这是非常严格的，要求版本正好是1.然而，未来发生了，并且发明了新版本的语法。现在，我要添加

friend struct NewGrammar<It, Skipper>;

使用旧语法并开始实现新语法，如果需要，它将慷慨地回归旧语法：

template <typename It, typename Skipper>
struct NewGrammar : qi::grammar<It, Skipper, std::string()>
{
    NewGrammar() : NewGrammar::base_type(mainrule)
    {
        using namespace qi;
        new_rule1 = int_(2); // support version 2 now
        new_start = omit [ "version" >> new_rule1 ] >> old.rule2; // note, no expectation point

        mainrule = new_start 
                 | old.mainrule;  // or fall back to version 1 grammar
    }
  private:
    OldGrammar<It, Skipper> old;
    qi::rule<It, Skipper, std::string()> new_start, mainrule;
    qi::rule<It, Skipper, int()>         new_rule1;
};

（我没有尝试过使用继承，但很有可能它也可以使用。）

让我们来试试这个宝贝：

template <template <typename It,typename Skipper> class Grammar>
bool test(std::string const& input)
{
    auto f(input.begin()), l(input.end());
    static const Grammar<std::string::const_iterator, qi::space_type> p;
    try {
        return qi::phrase_parse(f,l,p,qi::space) && (f == l); // require full input consumed
    } 
    catch(...) { return false; } // qi::expectation_failure<>
}

int main()
{
    assert(true  == test<OldGrammar>("version 1 woot"));
    assert(false == test<OldGrammar>("version 2 nope"));

    assert(true  == test<NewGrammar>("version 1 woot"));
    assert(true  == test<NewGrammar>("version 2 woot as well"));
}

显然所有测试都通过了： see it live on Coliru ¹希望这会有所帮助！

¹好吧，笨蛋。 Coliru今天编译速度太慢了。所以这是完整的测试程序：

#define BOOST_SPIRIT_DEBUG
#include <boost/spirit/include/qi.hpp>

namespace qi = boost::spirit::qi;

template <typename It, typename Skipper>
struct NewGrammar; // forward declare for friend declaration

template <typename It, typename Skipper>
struct OldGrammar : qi::grammar<It, Skipper, std::string()>
{
    friend struct NewGrammar<It, Skipper>; // NOTE

    OldGrammar() : OldGrammar::base_type(mainrule)
    {
        using namespace qi;
        rule1 = int_(1); // expect version 1
        rule2 = *char_;  // hopefully some interesting grammar
        mainrule = omit [ "version" > rule1 ] >> rule2;

        BOOST_SPIRIT_DEBUG_NODE(mainrule);
        BOOST_SPIRIT_DEBUG_NODE(rule1);
        BOOST_SPIRIT_DEBUG_NODE(rule2);
    }
  private:
    qi::rule<It, Skipper, std::string()> mainrule;
    qi::rule<It, Skipper, int()>         rule1;
    qi::rule<It, Skipper, std::string()> rule2;
};

template <typename It, typename Skipper>
struct NewGrammar : qi::grammar<It, Skipper, std::string()>
{
    NewGrammar() : NewGrammar::base_type(mainrule)
    {
        using namespace qi;
        new_rule1 = int_(2); // support version 2 now
        new_start = omit [ "version" >> new_rule1 ] >> old.rule2; // note, no expectation point

        mainrule = new_start 
                 | old.mainrule;  // or fall back to version 1 grammar

        BOOST_SPIRIT_DEBUG_NODE(new_start);
        BOOST_SPIRIT_DEBUG_NODE(mainrule);
        BOOST_SPIRIT_DEBUG_NODE(new_rule1);
    }
  private:
    OldGrammar<It, Skipper> old;
    qi::rule<It, Skipper, std::string()> new_start, mainrule;
    qi::rule<It, Skipper, int()>         new_rule1;
};

template <template <typename It,typename Skipper> class Grammar>
bool test(std::string const& input)
{
    auto f(input.begin()), l(input.end());
    static const Grammar<std::string::const_iterator, qi::space_type> p;
    try {
        return qi::phrase_parse(f,l,p,qi::space) && (f == l); // require full input consumed
    } 
    catch(...) { return false; } // qi::expectation_failure<>
}

int main()
{
    assert(true  == test<OldGrammar>("version 1 woot"));
    assert(false == test<OldGrammar>("version 2 nope"));

    assert(true  == test<NewGrammar>("version 1 woot"));
    assert(true  == test<NewGrammar>("version 2 woot as well"));
}

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