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在 boost::spirit 中，我添加了基于 example roman 的错误处理代码。

#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
#include <boost/spirit/include/phoenix_core.hpp>
#include <boost/spirit/include/phoenix_fusion.hpp>
#include <boost/spirit/include/phoenix_object.hpp>
#include <boost/foreach.hpp>

#include <iostream>
#include <fstream>
#include <string>
#include <vector>

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

template <typename Iterator>
struct roman : qi::grammar<Iterator>
{
  roman() : roman::base_type(start)
  {
    using qi::eps;
    using qi::lit;
    using qi::lexeme;
    using qi::_val;
    using qi::_1;
    using ascii::char_;

    // for on_error
    using qi::on_error;
    using qi::fail;
    using phoenix::construct;
    using phoenix::val;

    start = +(lit('M') )  >> "</>";

    on_error<fail>
    (
        start
      , std::cout
            << val("Error! Expecting ")
            // << _4                            // what failed?
            << val(" here: \"")
            // << construct<std::string>(_3, _2)   // iterators to error-pos, end
            << val("\"")
            << std::endl
    );
  }
  qi::rule<Iterator> start;
};

int
main()
{
    std::cout << "/////////////////////////////////////////////////////////\n\n";
    std::cout << "\t\tRoman Numerals Parser\n\n";
    std::cout << "/////////////////////////////////////////////////////////\n\n";
    std::cout << "Type a Roman Numeral ...or [q or Q] to quit\n\n";

    typedef std::string::const_iterator iterator_type;
    typedef roman<iterator_type> roman;

    roman roman_parser; // Our grammar

    std::string str;
    unsigned result;
    while (std::getline(std::cin, str))
    {
        if (str.empty() || str[0] == 'q' || str[0] == 'Q')
            break;

        std::string::const_iterator iter = str.begin();
        std::string::const_iterator end = str.end();
        //[tutorial_roman_grammar_parse
        bool r = parse(iter, end, roman_parser, result);

        if (r && iter == end)
        {
            std::cout << "-------------------------\n";
            std::cout << "Parsing succeeded\n";
            std::cout << "result = " << result << std::endl;
            std::cout << "-------------------------\n";
        }
        else
        {
            std::string rest(iter, end);
            std::cout << "-------------------------\n";
            std::cout << "Parsing failed\n";
            std::cout << "stopped at: \": " << rest << "\"\n";
            std::cout << "-------------------------\n";
        }
        //]
    }

    std::cout << "Bye... :-) \n\n";
    return 0;
}

我的问题是：

  1. “on_error”没有被触发，为什么？
  2. 我注释了“<< _4”，如果我想打印出失败的部分，该怎么做？
4

1 回答 1

7

三个步骤：

  1. 限定占位符：

    on_error<fail>(start, 
        std::cout
           << val("Error! Expecting ")
           << qi::_4
           << val(" here: \"")
           << construct<std::string>(qi::_3, qi::_2)
           << val("\"")
           << std::endl
    );

  2. 您还需要确保您有触发错误处理程序的期望点。

    start = eps > +(lit('M') ) >> "</>";

    参见例如Boost.Spirit.Qi - Errors at the beginning of a rule以获得解释

  3. （可选）命名您的规则

    start.name("start");

    使用 BOOST_SPIRIT_DEBUG_NODE(S) 是另一种隐式命名规则的方法。

在 Coliru 上看到它（在某些地方进行了清理和简化）

现在它打印（输入iv）：

Error! Expecting <sequence>"M""</>" here: 'iv'
Parsing failed
stopped at: 'iv'

完整代码

#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>

#include <iostream>
#include <fstream>

namespace qi  = boost::spirit::qi;
namespace phx = boost::phoenix;

template <typename Iterator>
struct roman : qi::grammar<Iterator>
{
    roman() : roman::base_type(start)
    {
        using namespace qi;

        start = eps > +lit('M') >> "</>";
        start.name("start");

        on_error<fail>(start, 
                phx::ref(std::cout)
                   << "Error! Expecting "
                   << qi::_4
                   << " here: '"
                   << phx::construct<std::string>(qi::_3, qi::_2)
                   << "'\n"
            );
    }
    qi::rule<Iterator> start;
};

int main()
{
    typedef std::string::const_iterator iterator_type;
    roman<iterator_type> roman_parser; // Our grammar

    std::string str;
    while (std::getline(std::cin, str))
    {
        if (str.empty() || str[0] == 'q' || str[0] == 'Q')
            break;

        iterator_type iter = str.begin(), end = str.end();
        unsigned result;
        bool r = parse(iter, end, roman_parser, result);

        if (r && iter == end)
        {
            std::cout << "Parsing succeeded\n";
            std::cout << "result = " << result << std::endl;
        }
        else
        {
            std::string rest(iter, end);
            std::cout << "Parsing failed\n";
            std::cout << "stopped at: '" << rest << "'\n";
        }
    }
}

除了评论：这是我一直在测试的东西 - 还没有完全让它工作，但是错误处理程序正在被调用并按照它应该的方式吃输入。也许它会有所帮助？

static auto const at_eol = (*_1 == '\r') || (*_1 == '\n');
static auto const at_eoi = (_1 == _2);

on_error<retry>(start, 
    (
        (phx::ref(std::cout) << "rule start: expecting " << _4 << " here: '" << escape_(_3, _2) << "'\n"),
        phx::while_ (!at_eoi && !at_eol) [ ++_1, phx::ref(std::cout) << "\nadvance to newline\n" ],
        phx::while_ (!at_eoi && at_eol)  [ ++_1, phx::ref(std::cout) << "\neat newline\n" ],
        phx::if_ (at_eoi)                [ _pass = fail ]
    )
);

另请参阅Importantmulti_pass<> 文档中的注释

于 2013-10-15T11:59:19.270 回答