在一个更简单的环境中,这表明这主要是类型推断的限制:
Derived parse_result;
bool ok = qi::phrase_parse(f, l, base_, qi::space, data);
当解析器暴露一个Base时,该方法将无法工作,但是您可以通过为模板实例化添加“类型提示”来修复它[1]:
bool ok = qi::phrase_parse(f, l, base_, qi::space, static_cast<Base&>(data));
完整演示在Coliru上实时演示
#include <algorithm>
#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
namespace qi = boost::spirit::qi;
namespace phoenix = boost::phoenix;
struct Base {
int x;
double y;
};
BOOST_FUSION_ADAPT_STRUCT(Base, (int,x)(double,y))
struct Derived : Base { };
int main()
{
typedef std::string::const_iterator It;
qi::rule<It, Base(), qi::space_type> base_ = qi::int_ >> qi::double_;
std::string const input = "1 3.14";
auto f(input.begin()), l(input.end());
Derived parse_result;
bool ok = qi::phrase_parse(f, l, base_, qi::space, static_cast<Base&>(parse_result));
if (ok)
{
std::cout << "Parsed: " << parse_result.x << " " << parse_result.y << "\n";
} else
{
std::cout << "Parse failed\n";
}
if (f != l)
{
std::cout << "Input remaining: '" << std::string(f,l) << "'\n";
}
}
或者
您可以通过显式传递可公开属性的引用给基本解析器/规则来避免混淆:
template <typename It, typename Skipper = qi::space_type>
struct derived_grammar : qi::grammar<It, Derived(), Skipper>
{
derived_grammar() : derived_grammar::base_type(start) {
base_ = qi::int_ >> qi::double_;
glue_ = base_ [ qi::_r1 = qi::_1 ];
start = "derived:" >> glue_(qi::_val);
}
private:
qi::rule<It, Derived(), Skipper> start;
qi::rule<It, void(Base&), Skipper> glue_;
qi::rule<It, Base(), Skipper> base_;
};
如果你坚持的话,可以使用
qi::attr_cast<Base, Base>
来避免使用
glue_
/
base_
分离 (但为了可读性,我不会这样做)。
以下是完整代码以供参考:
在Coliru上实时运行。
#define BOOST_SPIRIT_USE_PHOENIX_V3
#include <algorithm>
#include <boost/fusion/adapted/struct.hpp>
#include <boost/spirit/include/qi.hpp>
#include <boost/spirit/include/phoenix.hpp>
namespace qi = boost::spirit::qi;
namespace phx = boost::phoenix;
struct Base {
int x;
double y;
};
BOOST_FUSION_ADAPT_STRUCT(Base, (int,x)(double,y))
struct Derived : Base { };
template <typename It, typename Skipper = qi::space_type>
struct derived_grammar : qi::grammar<It, Derived(), Skipper>
{
derived_grammar() : derived_grammar::base_type(start) {
base_ = qi::int_ >> qi::double_;
glue_ = base_ [ qi::_r1 = qi::_1 ];
start = "derived:" >> glue_(qi::_val);
}
private:
qi::rule<It, Derived(), Skipper> start;
qi::rule<It, void(Base&), Skipper> glue_;
qi::rule<It, Base(), Skipper> base_;
};
int main()
{
typedef std::string::const_iterator It;
derived_grammar<It> g;
std::string const input = "derived:1 3.14";
auto f(input.begin()), l(input.end());
Derived parse_result;
bool ok = qi::phrase_parse(f, l, g, qi::space, parse_result);
if (ok)
{
std::cout << "Parsed: " << parse_result.x << " " << parse_result.y << "\n";
} else
{
std::cout << "Parse failed\n";
}
if (f != l)
{
std::cout << "Input remaining: '" << std::string(f,l) << "'\n";
}
}
[1] 此处指的是 qi::phrase_parse
函数模板的实例化过程,其中包括 函数模板 和 实例化。
static_cast
的那个),而不是立即使用phrase_parse
吗? - user2913094