你怎么用一个变量储存在一个提高精神封闭作为输入到提升精神环析器?

Question

我想用一分析的价值作为输入到一个循环分析器。

该法定义了一个标题，用于指定的(变量)的尺寸以下的串。例如，说以下的串被输入到一些分析器。

12 Test有效载荷

分析程序应该提取 12, 转换到一个 unsigned int 然后读十二字符。我可以定义的提升精神语汇编，但声明在提高精神码失败在运行时间。

#include <iostream>
#include <boost/spirit.hpp>

using namespace boost::spirit;

struct my_closure : public closure<my_closure, std::size_t> {
member1 size;
};

struct my_grammar : public grammar<my_grammar> {
template <typename ScannerT>
struct definition {
    typedef rule<ScannerT> rule_type;
    typedef rule<ScannerT, my_closure::context_t> closure_rule_type;

    closure_rule_type header;
    rule_type payload;
    rule_type top;

    definition(const my_grammar &self)
    {
        using namespace phoenix;
        header = uint_p[header.size = arg1];
        payload = repeat_p(header.size())[anychar_p][assign_a(self.result)];
        top = header >> str_p("\r\n") >> payload;
    }

    const rule_type &start() const { return top; }
};

my_grammar(std::string &p_) : result(p_) {}
std::string &result;
};

int
main(int argc, char **argv)
{
const std::string content = "12\r\nTest Payload";
std::string payload;
my_grammar g(payload);
if (!parse(content.begin(), content.end(), g).full) {
    std::cerr << "there was a parsing error!\n";
    return -1;
}
std::cout << "Payload: " << payload << std::endl;
return 0;
}

是否可以告诉精神封闭的变量进行评估，懒洋洋地?是这种行为，支持通过提高精神？

Answer 1

这是与所提供的新气解析器灵2.下面的代码段提供了一个完整的例子，大多作品容易得多。一个意想不到的字符被插入到最终的结果。

#include <iostream>
#include <string>

#include <boost/version.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/qi_repeat.hpp>
#include <boost/spirit/include/qi_grammar.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>

using boost::spirit::qi::repeat;
using boost::spirit::qi::uint_;
using boost::spirit::ascii::char_;
using boost::spirit::ascii::alpha;
using boost::spirit::qi::_1;
namespace phx = boost::phoenix;
namespace qi = boost::spirit::qi;

template <typename P, typename T>
void test_parser_attr(
    char const* input, P const& p, T& attr, bool full_match = true)
{
    using boost::spirit::qi::parse;

    char const* f(input);
    char const* l(f + strlen(f));
    if (parse(f, l, p, attr) && (!full_match || (f == l)))
        std::cout << "ok" << std::endl;
    else
        std::cout << "fail" << std::endl;
}

static void
straight_forward()
{
    std::string str;
    int n;
    test_parser_attr("12\r\nTest Payload",
                     uint_[phx::ref(n) = _1] >> "\r\n" >> repeat(phx::ref(n))[char_],
                     str);
    std::cout << "str.length() == " << str.length() << std::endl;
    std::cout << n << "," << str << std::endl;  // will print "12,Test Payload"
}

template <typename P, typename T>
void
test_phrase_parser(char const* input, P const& p,
                   T& attr, bool full_match = true)
{
    using boost::spirit::qi::phrase_parse;
    using boost::spirit::qi::ascii::space;

    char const* f(input);
    char const* l(f + strlen(f));
    if (phrase_parse(f, l, p, space, attr) && (!full_match || (f == l)))
        std::cout << "ok" << std::endl;
    else
        std::cout << "fail" << std::endl;
}

template <typename Iterator>
struct test_grammar
    : qi::grammar<Iterator, std::string(), qi::locals<unsigned> > {

    test_grammar()
        : test_grammar::base_type(my_rule)
    {
        using boost::spirit::qi::_a;
        my_rule %= uint_[_a = _1] >> "\r\n" >> repeat(_a)[char_];
    }

    qi::rule<Iterator, std::string(), qi::locals<unsigned> > my_rule;
};

static void
with_grammar_local_variable()
{
    std::string str;
    test_phrase_parser("12\r\nTest Payload", test_grammar<const char*>(), str);
    std::cout << str << std::endl;  // will print "Test Payload"
}

int
main(int argc, char **argv)
{
    std::cout << "boost version: " << BOOST_LIB_VERSION << std::endl;

    straight_forward();
    with_grammar_local_variable();

    return 0;
}

Answer 2

你在找什么是lazy_p，检查这里的例子: http://www.boost.org/doc/libs/1_35_0/libs/spirit/doc/the_lazy_parser.html