用模板化类重载lexical_Cast

Question

我试图扩展lexical_cast以处理字符串->cv：：Point转换，代码如下：

#include <iostream>
#include <opencv2/opencv.hpp>
#include <boost/lexical_cast.hpp>
#include <boost/algorithm/string/classification.hpp>
#include <boost/algorithm/string/split.hpp>

namespace boost {
  template<>
    cv::Point2f lexical_cast(const std::string &str) {
      std::vector<std::string> parts;
      boost::split(parts, str, boost::is_any_of(","));
      cv::Point2f R;
      R.x = boost::lexical_cast<float>(parts[0]);
      R.y = boost::lexical_cast<float>(parts[1]);
      return R;
    }
}

int main(int argc, char **argv) {
  auto p = boost::lexical_cast<cv::Point2f>(std::string("1,2"));
  std::cout << "p = " << p << std::endl;
  return 0;
}

效果很好..。然而，cv::Point2f实际上是cv::Point_<T>，其中T可以是int、float、double等。无论如何，我都找不到将这个模板化的arg公开给lexical_cast，这样我就可以有一个能够处理所有cv::Point_<T>类型的lexical_cast函数。

Answer 1

template <typename T>
struct point_type {};

template <typename T>
struct point_type<cv::Point_<T>> { using type = T; };

namespace boost {
  template <typename T, typename U = typename point_type<T>::type>
    T lexical_cast(const std::string &str)
    {
      std::vector<std::string> parts;
      boost::split(parts, str, boost::is_any_of(","));
      T R;
      R.x = boost::lexical_cast<U>(parts[0]);
      R.y = boost::lexical_cast<U>(parts[1]);
      return R;
    }
}

DEMO

前面的一个更复杂的解决方案，如果您不喜欢lexical_cast的第二个隐式模板参数

#include <type_traits>

template <typename T>
struct is_point : std::false_type {};

template <typename T>
struct is_point<cv::Point_<T>> : std::true_type {};

template <typename T>
struct point_type;

template <typename T>
struct point_type<cv::Point_<T>> { using type = T; };

namespace boost {
  template <typename T>
    auto lexical_cast(const std::string &str)
      -> typename std::enable_if<is_point<T>::value, T>::type
    {
      std::vector<std::string> parts;
      boost::split(parts, str, boost::is_any_of(","));
      using U = typename point_type<T>::type;
      T R;
      R.x = boost::lexical_cast<U>(parts[0]);
      R.y = boost::lexical_cast<U>(parts[1]);
      return R;
    }
}

DEMO 2