Question

是否可以从Spirit解析器的语义操作中检索词法分析器中定义的标记的ID，如：token_def<> tok;。

我尝试做的是使用每个运算符的标记ID（如+， - ，*等），并从解析器语义操作中检索该运算符的相应名称（例如，添加，减去，时间等。）

据我了解，在以下的制作中：

toks.symbol＆gt;＆gt; toks.plus＆gt;＆gt; toks.symbol;

如果toks.plus是token_def<> plus;或类型toks.symbol，则_1将引用第一个toks.symbol，_2将引用第二个{{1}}。真的吗？如果是这样，我如何访问中间令牌（只是为了检索ID）？

谢谢！

Answer 1

您可以使用内置的延迟占位符lex::_tokenid，请参阅文档：

Lexer Semantic Actions Using Phoenix

我调整了second word-count sample from the tutorials来动态打印令牌ID：

#include <boost/config/warning_disable.hpp>
#include <boost/spirit/include/lex_lexertl.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/include/phoenix_statement.hpp>
#include <boost/spirit/include/phoenix_algorithm.hpp>
#include <boost/spirit/include/phoenix_core.hpp>

#include <iostream>
#include <string>

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

struct distance_func
{
    template <typename Iterator1, typename Iterator2>
    struct result : boost::iterator_difference<Iterator1> {};

    template <typename Iterator1, typename Iterator2>
    typename result<Iterator1, Iterator2>::type 
    operator()(Iterator1& begin, Iterator2& end) const
    {
        return std::distance(begin, end);
    }
};
boost::phoenix::function<distance_func> const distance = distance_func();

template <typename Lexer>
struct word_count_tokens : lex::lexer<Lexer>
{
    word_count_tokens()
      : c(0), w(0), l(0)
      , word("[^ \t\n]+")     // define tokens
      , eol("\n")
      , any(".")
    {
        using boost::spirit::lex::_start;
        using boost::spirit::lex::_end;
        using boost::spirit::lex::_tokenid;
        using boost::phoenix::ref;

        // associate tokens with the lexer
        this->self 
            =   word  [++ref(w), ref(c) += distance(_start, _end), phx::ref(std::cout) << _tokenid << ";" ]
            |   eol   [++ref(c), ++ref(l), phx::ref(std::cout) << _tokenid << ";" ] 
            |   any   [++ref(c), phx::ref(std::cout) << _tokenid << ";" ]
            ;
    }

    std::size_t c, w, l;
    lex::token_def<> word, eol, any;
};

///////////////////////////////////////////////////////////////////////////////
int main(int argc, char* argv[])
{
    typedef 
        lex::lexertl::token<char const*, lex::omit, boost::mpl::false_> 
        token_type;

    typedef lex::lexertl::actor_lexer<token_type> lexer_type;

    word_count_tokens<lexer_type> word_count_lexer;

    std::string str ("the lazy moon jumped over the brazen mold");
    char const* first = str.c_str();
    char const* last = &first[str.size()];

    lexer_type::iterator_type iter = word_count_lexer.begin(first, last);
    lexer_type::iterator_type end = word_count_lexer.end();

    while (iter != end && token_is_valid(*iter))
        ++iter;

    if (iter == end) {
        std::cout << "\nlines: " << word_count_lexer.l 
            << ", words: " << word_count_lexer.w 
            << ", characters: " << word_count_lexer.c 
            << "\n";
    }
    else {
        std::string rest(first, last);
        std::cout << "Lexical analysis failed\n" << "stopped at: \"" 
            << rest << "\"\n";
    }
    return 0;
}

输出：

65536;65538;65536;65538;65536;65538;65536;65538;65536;65538;65536;65538;65536;65538;65536;
lines: 0, words: 8, characters: 41

检索token_def＆lt;＆gt;的ID

1 个答案: