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我想根据以前解析的值定义一个规则,即输入字符串具有以下结构:D <double number>I <integer number>。无论第一个读取字符是D还是,我都将其保存在本地布尔变量中I。完整的代码是:

#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <string>

namespace qi = boost::spirit::qi;
namespace spirit = boost::spirit;
namespace ascii = boost::spirit::ascii;
using boost::phoenix::ref;

template <typename Iterator>
struct x_grammar : public qi::grammar<Iterator, std::string(), ascii::space_type>
{
public:
    x_grammar() : x_grammar::base_type(start_rule, "x_grammar")
    {
        using namespace qi;
        bool is_int = false;
        start_rule = lit("I")[ref(is_int) = true] | lit("D")[ref(is_int) = false] > digit_rule;
        if(ref(is_int)()) {
            digit_rule = int_[std::cout << "int " << _1 << ".\n"];
        } else {
            digit_rule = double_[std::cout << "double " << _1 << ".\n"];
        }
    }
private:
    qi::rule<Iterator, std::string(), ascii::space_type> start_rule;
    qi::rule<Iterator, std::string(), ascii::space_type> digit_rule;
};

int main()
{
    typedef std::string::const_iterator iter;
    std::string storage("I 5");
    iter it_begin(storage.begin());
    iter it_end(storage.end());
    std::string read_data;
    using boost::spirit::ascii::space;
    x_grammar<iter> g;
    try {
        bool r = qi::phrase_parse(it_begin, it_end, g, space, read_data);
        if(r) {
            std::cout << "Pass!\n";
        } else {
            std::cout << "Fail!\n";
        }
    } catch (const qi::expectation_failure<iter>& x) {
        std::cout << "Fail!\n";
    }
    return 0;
}                                                                                                                                                    

输出是:double Pass!!!它既不识别if语句,也不打印解析后的数字!

注意:我知道还有其他直接的方法可以解析上面的示例。我必须解析的实际字符串看起来很复杂,这个例子只是说明了我想要实现的目标。总体目标是使用局部变量并基于这些变量定义其他规则。

我使用了 4.6.1 和 Boost 1.55 版本。

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1 回答 1

2
    if(ref(is_int)()) {

在这里,您可以评估施工期间的状况。这不是它的工作方式。该规则将始终采用相同的分支。

相反,请查看 Nabialek 技巧:http ://boost-spirit.com/home/articles/qi-example/nabialek-trick/

这是应用于您的样本Live On Coliru的完整 Nabialek 技巧:

  1. 你需要制作std::cout << "int" 懒惰的演员(至少包装phx::ref(std::cout)phx::val("int")作为凤凰演员)

  2. 您仍然没有使用属性传播 ( std::string()),因为它在存在语义操作的情况下被禁用(请参阅上一个答案)。不过,您可以从 Nabialek 子规则传播值:

#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <string>

namespace qi = boost::spirit::qi;
namespace spirit = boost::spirit;
namespace ascii = boost::spirit::ascii;
namespace phx = boost::phoenix;
using boost::phoenix::ref;


template <typename Iterator>
struct x_grammar : public qi::grammar<Iterator, ascii::space_type, qi::locals<qi::rule<Iterator, ascii::space_type>*> >
{
public:
    x_grammar() : x_grammar::base_type(start_rule, "x_grammar")
    {
        using namespace qi;

        int_rule = int_   [std::cout << phx::val("int ") << _1 << ".\n"];
        dbl_rule = double_[std::cout << phx::val("double ") << _1 << ".\n"];
        subrules.add
            ("I", &int_rule)
            ("D", &dbl_rule);

        start_rule = subrules[_a = _1] >> lazy(*_a);
    }
private:
    typedef qi::rule<Iterator, ascii::space_type> subrule;

    qi::symbols<char, subrule*> subrules;
    qi::rule<Iterator, ascii::space_type, qi::locals<subrule*> > start_rule;
    qi::rule<Iterator, ascii::space_type> int_rule, dbl_rule;
};

int main()
{
    typedef std::string::const_iterator iter;
    std::string storage("I 5");
    iter it_begin(storage.begin());
    iter it_end(storage.end());
    using boost::spirit::ascii::space;
    x_grammar<iter> g;
    try {
        bool r = qi::phrase_parse(it_begin, it_end, g, space);
        if (r) {
            std::cout << "Pass!\n";
        }
        else {
            std::cout << "Fail!\n";
        }
    }
    catch (const qi::expectation_failure<iter>&) {
        std::cout << "Fail!\n";
    }
    return 0;
}
于 2014-05-20T15:39:31.913 回答