Boost:spirit 解析为结构并重用其中的一部分

Boost:spirit Parsing into structure and reusing parts of it

我必须在句子中找到变量并用它们的值替换它们。变量可以写成不同的形式,例如 $varName 或 $(varName)。

我想要一个 struct VariableHolder 以便轻松访问两者:

struct VariableHolder
{
    string name;    // contains "varName"
    string fromFile;  // contains "$(varName)" or "$varName"
    void setName(ustring n) { name = n; }
}

显然,我想避免进行多次传递并调用多个解析器。 到目前为止我所拥有的是:

BOOST_FUSION_ADAPT_STRUCT(VariableHolder,
(ustring, fromFile)
)
// variableName is another parser that returns a string
qi::rule<Iterator, VariableHolder()> variable %=
                (qi::char_("$")
                    >> (variableName[phoenix::bind(&VariableHolder::setName, qi::_val, qi::_1)]
                        | (qi::char_("(")
                            >> variableName[phoenix::bind(&VariableHolder::setName, qi::_val, qi::_1)]
                            >> qi::char_(")")))
                    );

这是行不通的。名称设置正确,但 fromFile 变量只包含一个“$”,没有任何其他内容。

所以,我想问你们这些好人:

  1. 我的想法是只用 BOOST_FUSION_ADAPT_STRUCT 调整部分结构,并用语义动作填充其余部分。愚蠢的想法,还是我做错了?
  2. 有没有一种方法可以绑定语义操作并仍然获得输出?喜欢

    char_[doSomething] // 这可以同时调用 doSomething 并解析一个字符吗?

解决问题:

  1. My idea was to only adapt part of the structure with BOOST_FUSION_ADAPT_STRUCT, and fill the rest with semantic actions. Stupid idea, or am I just doing it wrong?

并非不可想象。不是我的推荐(参见 Boost Spirit: "Semantic actions are evil"?)。但是,是的,你做错了:

如果您不希望它成为公开属性的一部分,您需要 lit("$") 而不是 char_("$")。其实'$'这里就可以了

  1. Is there a way to bind a semantic action and still get the output? Like

    char_[doSomething] // Can this both call doSomething, and parse a char?
    

是的。您现在正在这样做,并且它之所以有效,是因为您使用了 operator%= 而不是 operator=(请参阅文档:http://www.boost.org/doc/libs/1_61_0/libs/spirit/doc/html/spirit/qi/reference/nonterminal/rule.html#spirit.qi.reference.nonterminal.rule.expression_semantics)。

但是,您似乎真的在尝试使用相同的输入两次(原始为 fromFile 和 "cooked" 为 name?)适得其反,因为自动规则属性propagation also 用你想要的值覆盖 name fromFile.

这里唯一快速的出路是只使用 SA。不过,我建议让 VariableHolder 的构造函数负责细节。

Side note: it looks a bit as if the optional parentheses suggest an expression grammar. If so, make that explicit in the grammar, instead of hardcoding a special case in the rule for a variableName. If not, carry on :)

这是一个尝试修复:

Live On Coliru

#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>

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

struct VariableHolder {
    std::string name;     // contains "varName"
    std::string fromFile; // contains "$(varName)" or "$varName"
};

template <typename It, typename Skipper = qi::ascii::space_type> struct P : qi::grammar<It, VariableHolder(), Skipper> {
    P() : P::base_type(start) {
        auto _name     = phoenix::bind(&VariableHolder::name, qi::_val);
        auto _fromFile = phoenix::bind(&VariableHolder::fromFile, qi::_val);

        variableName = qi::alpha >> +qi::alnum;
        variable = '$' >> (variableName | '(' >> variableName >> ')');

        start = qi::as_string [ qi::raw [ 
                variable [ _name = qi::_1 ]] 
            ] [ _fromFile = qi::_1 ];

        BOOST_SPIRIT_DEBUG_NODES((start)(variable)(variableName))
    }

  private:
    qi::rule<It, std::string(), Skipper> variable;
    qi::rule<It, VariableHolder(), Skipper> start;
    // lexemes
    qi::rule<It, std::string()> variableName;
};

int main() {
    using It = std::string::const_iterator;
    P<It> const p;

    for (std::string const input : { 
            "$foo1",
            "$(bar2)" 
        }) 
    {
        It f = input.begin(), l = input.end();
        VariableHolder data;

        bool ok = qi::phrase_parse(f, l, p, qi::ascii::space, data);

        if (ok) {
            std::cout << "Parse success: " << data.name << " (source: '" << data.fromFile << "')\n";
        } else {
            std::cout << "Parse failure ('" << input << "')\n";
        }

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

版画

Parse success: foo1 (source: '$foo1')
Parse success: bar2 (source: '$(bar2)')