Question

假设我要解析以下字符串：

“1.2,2.0,3.9”

当我为它应用以下解析器时：

struct DataStruct
{
    double n1, n2, n3;
};

BOOST_FUSION_ADAPT_STRUCT(DataStruct, (double, n1)(double, n2)(double, n3))

qi::rule<std::string::iterator, DataStruct()> data_ =
                                                      qi::double_ >> ','
                                                   >> qi::double_ >> ','
                                                   >> qi::double_;

auto str = "1.2, 2.0, 3.9";
auto it - str.begin();
if (qi::parse(it, str.end(), data_, res))
{
    std::cout << "parse completed" << std::endl;
}

一切都还可以，但是当我想在我的字符串中不是一些double时，我可以得到“null”（即“1.2，null，3.9”）我想在DataStruct中将0值赋给适当的double值。有办法做到这一点吗？

Answer 1

通常的诀窍是使用qi::attr的替代方案：

rule_def = parser_expression | qi::attr(default_value);

在你的情况下，也许：

reader_ = qi::double_ | qi::lit("null") >> qi::attr(0);

演示

<强> Live On Coliru

#include <boost/spirit/include/qi.hpp>
#include <boost/fusion/include/adapt_struct.hpp>
struct DataStruct { double n1, n2, n3; };

BOOST_FUSION_ADAPT_STRUCT(DataStruct, n1, n2, n3)

namespace qi = boost::spirit::qi;

int main() {
    using Iterator = typename std::string::const_iterator;

    qi::rule<Iterator, double()> reader_   = qi::double_ | qi::lit("null") >> qi::attr(0);
    qi::rule<Iterator, DataStruct()> data_ = reader_ >> ',' >> reader_ >> ',' >> reader_;

    DataStruct res;
    auto const str = std::string("1.2,null,3.9");
    Iterator start = str.begin(), end = str.end();

    if (qi::parse(start, end, data_ >> qi::eoi, res)) {
        std::cout << "parsed: " << boost::fusion::as_vector(res) << "\n";
    }
    else {
        std::cout << "parse failed\n";
    }
}

打印

parsed: (1.2 0 3.9)

请注意审核更改（不使用命名空间，请检查eoi）。

Answer 2

您可以使用以下代码：

#include <boost/fusion/include/adapt_struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
#include <boost/spirit/include/phoenix_operator.hpp>
struct DataStruct
{
    double n1, n2, n3;
};

BOOST_FUSION_ADAPT_STRUCT(DataStruct, (double, n1)(double, n2)(double, n3))

int main() {
    using namespace boost::spirit;

    using Iterator = typename std::string::iterator;

    qi::rule<Iterator, double()> reader_ = (qi::double_[_val = _1] | (lexeme["null"])[_val = 0.]);

    qi::rule<Iterator, DataStruct()> data_ = reader_ >> ',' >> reader_ >> ',' >> reader_;

    try {
        DataStruct res;
        auto str = std::string("1.2,null,3.9");
        auto start = str.begin();
        auto end = str.end();

        bool ok = qi::parse(start, end, data_, res);
        if (ok) {
            std::cout << "parse completed" << std::endl;
        }
        else {
            std::cout << "parse failed" << std::endl;
        }
    }
    catch (const qi::expectation_failure<std::string::iterator>& except) {
        std::cout << except.what();
    }

    std::cout << std::endl;
}

使用boost spirit将默认值分配给变量

2 个答案:

演示