如何以激励精神构建默认值语法

您要求的解析器非常复杂，因为它必须解决几个任务:

• 处理末尾缺失的元素
• 处理"2*"语法作为末尾缺失元素的替换
• 不仅正确解析所有有效的输入，而且用匹配的值填充给定的数据结构

• 为缺少的元素提供默认值:

  qi::int_ | qi::attr(180)
  

  。匹配整数或使用默认值180

• 为"2*"语法提供所有剩余的值(如@vines建议的):

  "2*" >> qi::attr(attr2)
  

  。如果2*匹配，则使用默认值attr2(即fusion::vector)。

总的来说，我想出了这个解决方案，它似乎解析并返回默认值很好(即使它看起来非常复杂):

#include <string>
#include <iostream>
#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/include/vector.hpp>
int main()
{
    namespace qi = boost::spirit::qi;
    namespace fusion = boost::fusion;
    // the attribute passed to the parser has to match (in structure) the 
    // parser, requiring to create nested fusion::vector's
    typedef fusion::vector<std::string, int>              attribute1_type;
    typedef fusion::vector<int, attribute1_type>          attribute2_type;
    typedef fusion::vector<int, attribute2_type>          attribute3_type;
    // overall attribute type
    typedef fusion::vector<std::string, attribute3_type>  attribute_type;
    // initialize attributes with default values
    attribute1_type attr1("male", 32);
    attribute2_type attr2(80, attr1);
    attribute3_type attr3(180, attr2);
    qi::rule<std::string::iterator, std::string()> quoted_string =
        "'" >> *~qi::char_("'") >> "'";
    qi::rule<std::string::iterator, attribute_type(), qi::space_type> data =
        qi::lit("Person") >> "{" 
            >>  quoted_string 
            >> -(   ("4*" >> qi::attr(attr3))
                |   (qi::int_ | qi::attr(180))
                    >> -(   ("3*" >> qi::attr(attr2))
                        |   (qi::int_ | qi::attr(80))
                            >> -(   ("2*" >> qi::attr(attr1))
                                |   (quoted_string | qi::attr("male"))
                                    >> -(   "1*"  
                                        |   qi::int_ 
                                        |   qi::attr(32)
                                        )
                                )
                        )
                )
        >> "}";
    std::string in1 = "Personn{ 'Tom' 188 80 'male' 32 }";
    attribute_type fullattr1;
    if (qi::phrase_parse(in1.begin(), in1.end(), data, qi::space, fullattr1))
        std::cout << fullattr1 << std::endl;
    std::string in2 = "Personn{ 'Tom' 188 80 'male' }";
    attribute_type fullattr2;
    if (qi::phrase_parse(in2.begin(), in2.end(), data, qi::space, fullattr2))
        std::cout << fullattr2 << std::endl;
    std::string in3 = "Personn{ 'Tom' 188 3* }";
    attribute_type fullattr3;
    if (qi::phrase_parse(in3.begin(), in3.end(), data, qi::space, fullattr3))
        std::cout << fullattr3 << std::endl;
    return 0;
}

将规则拆分为单独的规则(如@vines所建议的)将需要对输入进行多次解析，这就是为什么我使用这种序列和替代的嵌套结构。

我刚刚提出了一个通用的解决方案，虽然它有点复杂=)
它处理"大括号"和多个任意跳过说明符。这是:

#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/fusion/include/io.hpp>
#include <iostream>
#include <string>

namespace qi = boost::spirit::qi;
namespace ph = boost::phoenix;
struct numbers { int i1, i2, i3, i4; };
BOOST_FUSION_ADAPT_STRUCT
(numbers,
    (int, i1)
    (int, i2)
    (int, i3)
    (int, i4)
)
template <typename Iterator, typename Skipper>
struct Grammar : public qi::grammar <Iterator, numbers(), Skipper>
{
    Grammar() : Grammar::base_type(start, "numbers")
    {
    using qi::int_;
    // This rule resets the skip counter:
    init_skip = qi::eps[ph::ref(skp) = 0];
    // This rule parses the skip directive ("n*") and sets the skip counter:
    skip_spec = qi::omit[ (qi::lexeme[ int_ >> "*" ])[ph::ref(skp) = qi::_1] ];
    // This rule checks if we should skip the field, and if so, decrements
    // the skip counter and returns the value given to it (the default one).
    // If not, it tries to parse the int.
    // If int fails to parse, the rule resorts the default value again,
    // thus handling the "premature brace" case.
    int_dflt %= qi::eps(ph::ref(skp) > 0)[--ph::ref(skp)] >> qi::attr(qi::_r1) | int_ | qi::attr(qi::_r1);
    // And this is the grammar:
    start %= init_skip >>
             "{" >> -skip_spec >> int_dflt(-1)
                 >> -skip_spec >> int_dflt(-1)
                 >> -skip_spec >> int_dflt(-1)
                 >> -skip_spec >> int_dflt(-1)
                 >> "}";
    }
    // the skip counter itself:
    int skp;
    qi::rule <Iterator, numbers(), Skipper> start;
    qi::rule <Iterator, Skipper> skip_spec, init_skip;
    qi::rule <Iterator, int(int), Skipper> int_dflt;
};


int main (int argc, char **argv)
{
    using std::cout;
    using std::endl;
    std::string s = argv[1];
    numbers result;
    std::string::iterator ib = s.begin();
    std::string::iterator ie = s.end();
    bool r = qi::phrase_parse(ib, ie, Grammar<std::string::iterator, qi::space_type>(), qi::space, result );
    if (r && ib == ie)
    {
        cout << boost::fusion::tuple_open('[');
        cout << boost::fusion::tuple_close(']');
        cout << boost::fusion::tuple_delimiter(", ");
        cout << "Parsing succeededn";
        cout << "got: " << boost::fusion::as_vector(result) << endl;
    }
    else
    {
        cout << "Parsing failedn";
        cout << "err: " << std::string(ib, ie) << endl;
    }
    return 0;
}

PS:请注意，Skipper模板参数与字段跳过无关—它只是语法使用的空白跳过解析器的类型。

我首先想到的是:

如果你的结构没有太多的成员，你可以用某种语法来描述*n，比如:

struct_full = "{" >> a >> b >> c >> "}";
struct_reduced_1 = "{" >> a >> b >> "1*" >> attr(c_default) >> "}"
struct_reduced_2 = "{" >> a >> "2*" >> attr(b_default) >> attr(c_default) >> "}";
struct_reduced_3 = "{" >> "3*" >> attr(a_default) >> attr(b_default) >> attr(c_default) >> "}";

当然，这不是最美丽的方式…