Question

我一直在尝试使用Boost Spirit解析一个字符串，如下所示：

integer_count int1 int2 int3 ... intN

其中N是integer_count。例如，

5 1 2 3 4 5

代码如下：

#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";
    }   
}

输出结果为：

repeat_count == 0
Pass!

这意味着

ref(repeat_count) = _1

从未被调用过。那么有没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解析？

Answer 1

我认为你的意思是

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

此外，当存在语义动作时，对规则禁用自动属性传播。您可以使用%=强制进行属性传播

这是一个简单的固定版本 Live On 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!

注意显然，repeat_count == 0 总是打印，因为它是在语法构造过程中执行的，而不是解析。

未调用提升精神语义动作

1 个答案: