使用std :: variant增强boost :: mpi和boost:serialization

时间:2019-05-24 10:00:03

标签: c++17 boost-serialization boost-mpi std-variant

c ++ 17引入了新类型std::variant。是否可以定义一个序列化例程,以便与std::variant一起使用boost::mpi

考虑一个简单的程序

#include <variant>
#include <boost/mpi.hpp>
#include <boost/serialization/string.hpp>

namespace mpi = boost::mpi;

class A {
  friend class boost::serialization::access;

  template <class Archive>
  void serialize(Archive& ar, const unsigned int version)
  {
    ar & x;
    ar & y;
  }

public:
  int x, y;
};

class B {
  friend class boost::serialization::access;

  template <class Archive>
  void serialize(Archive& ar, const unsigned int version)
  {
    ar & z;
  }

public:
  int z;
};

int main()
{
  mpi::environment env;
  mpi::communicator world;

  std::variant<A, B> v;
  if (world.rank() == 0)
    v = B{1};
  mpi::broadcast(world, v, 0);

  return 0;
}

它不会编译给出错误

error: ‘class std::variant<A, B>’ has no member named ‘serialize’

一个人如何为serialize定义一个std::variant成员?请注意,上面示例中的AB类型都具有正确定义的序列化成员。

0 个答案:

没有答案