提神词及其属性

我正在使用一个跳过空白的解析器。有一次,我不想跳过,所以我想用 qi::lexeme lexeme 处理好了吗?

下面是我想做的一个例子:

#include <iostream>
#include <iomanip>
#include <string>
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/include/vector.hpp>

namespace qi = boost::spirit::qi;
namespace fu = boost::fusion;

struct printer_type
{
    void operator() (int i) const
    {
        std::cout << i << ' ';
    }

    void operator() (std::string s) const
    {
        std::cout << '"' << s << '"' << ' ';
    }

} printer;

int main() {
    for (std::string str : { "1foo 13", "42 bar 13", "13cheese 8", "101pencil13" }) {
        auto iter = str.begin(), end = str.end();

        qi::rule<std::string::iterator, qi::blank_type, fu::vector<int, std::string, int>()> parser = qi::int_ >> +qi::alpha >> qi::int_;

        fu::vector<int, std::string, int> result;
        bool r = qi::phrase_parse(iter, end, parser, qi::blank, result);

        std::cout << " --- " << std::quoted(str) << " --- ";
        if (r) {
            std::cout << "parse succeeded: ";
            fu::for_each(result, printer);
            std::cout << '\n';
        } else {
            std::cout << "parse failed.\n";
        }

        if (iter != end) {
            std::cout << " Remaining unparsed: " << std::string(iter, str.end()) << '\n';
        }
    }
}

注意这一行:

qi::rule<std::string::iterator, qi::blank_type, fu::vector<int, std::string, int>()> parser = 
                      qi::int_ >> +qi::alpha >> qi::int_;

好的,我们需要一个int,然后是一个字符串,然后是一个int。但是,我不想跳过第一个int和字符串之间的空白,这里不能有空格。如果我使用lexeme,合成的属性就会混乱。

无障碍跑步 词素 给出以下结果:

--- "1foo 13" --- parse succeeded: 1 "foo" 13 
 --- "42 bar 13" --- parse succeeded: 42 "bar" 13 
 --- "13cheese 8" --- parse succeeded: 13 "cheese" 8 
 --- "101pencil13" --- parse succeeded: 101 "pencil" 13 

所以一切都很好,这很好。然而,第二个例子( 42 bar 13 词素 在第一个int和字符串周围( qi::lexeme[qi::int_ >> +qi::alpha] >> qi::int_;

" 0  "1foo 13" --- parse succeeded: 1 "
 --- "42 bar 13" --- parse failed.
 Remaining unparsed: 42 bar 13
 --- "13cheese 8" --- parse succeeded: 13 " 0 
" 0  "101pencil13" --- parse succeeded: 101 "

旁白:我想略去 词素 并定义一个不跳过的子规则。在这种情况下,如何指定属性?

然后子规则具有 fusion::vector<int, std::string>() fusion::vector<int, std::string, int>() fusion::vector<fusion::vector<int, std::string>, int>() (反正也不会编译)。