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Question

我一直在尝试使用Boost精灵解析字符串，如下所示：

integer_count int1 int2 int3 ... intN

其中N是integer_count。例如,

5 1 2 3 4 5

守则如下：

#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; 
        int repeat_count = 0;
        start_rule = int_[ref(repeat_count) = _1] > repeater > *char_;
        std::cout << "repeat_count == " << repeat_count << std::endl;
        repeater = repeat(repeat_count)[int_[std::cout << _1 << ".\n"]];
    }   
private:
    qi::rule<Iterator, std::string(), ascii::space_type> start_rule;
    qi::rule<Iterator, std::string(), ascii::space_type> repeater;
};  

int main()
{   
    typedef std::string::const_iterator iter;
    std::string storage("5 1 2 3 4 5 garbage");
    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";
    }   
}   

产出如下：

repeat_count == 0
Pass!

这意味着

ref(repeat_count) = _1

从未被提起过。那么，是否有办法将整数运行时读入局部变量并使用该值进行进一步的解析？

Answer 1

我想你是说

repeat(boost::phoenix::ref(repeat_count))

此外，当有语义操作时，规则将禁用自动属性传播。您可以使用%=强制属性传播。

下面是一个简单的固定版本住在Coliru

#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, ascii::space_type>
{   
public:
    x_grammar() : x_grammar::base_type(start_rule, "x_grammar")
    {   
        using namespace qi; 
        int repeat_count = 0;
        start_rule = int_[ref(repeat_count) = _1] > repeater > *char_;
        std::cout << "repeat_count == " << repeat_count << std::endl;
        repeater = repeat(ref(repeat_count))[int_[std::cout << _1 << ".\n"]];
    }   
private:
    qi::rule<Iterator, ascii::space_type> start_rule;
    qi::rule<Iterator, ascii::space_type> repeater;
};  

int main()
{   
    typedef std::string::const_iterator iter;
    std::string storage("5 1 2 3 4 5 garbage");
    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>& x) {
        std::cout << "Fail!\n";
    }   
}   

打印

repeat_count == 0
1.
2.
3.
4.
5.
Pass!

显然，注意到repeat_count == 0是如何打印的，因为它是在语法构造过程中执行的，而不是解析过程中执行的。