c-使用Boost Spirit将默认值分配给变量

假设我要解析以下字符串：

“ 1.2、2.0、3.9”

当我为其应用以下解析器时：

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;
}

一切正常,但是当我假设我的字符串不是某个双精度数时,我可以得到“ null”(即“ 1.2,null,3.9”),我想将0值分配给DataStruct中适当的double值.有什么办法吗？

解决方法:

通常的技巧是使用qi :: attr的替代方法：

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

就您而言,也许：

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

演示

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)

注意检查更改(不要使用名称空间,请检查eoi).

标签：c,boost,boost-spirit
来源： https://codeday.me/bug/20191011/1893377.html