Question

如何序列化arma::Col？下面是MWE和错误输出。

MWE：

#include <boost/mpi/environment.hpp>
#include <boost/mpi/communicator.hpp>
#include <iostream>
#include "armadillo"

namespace mpi = boost::mpi;

struct S
{   
    int i;
    arma::Col<double>::fixed<3> cvector;

    friend class boost::serialization::access;

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

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

    S s;

    if (world.rank() == 0)
    {
        s.cvector[0] = 2;
        s.cvector[1] = 2;
        world.send(1, 0, s);
    }
    else
    {
        world.recv(0, 0, s);
        std::cout << s.cvector[0] << std::endl;
        std::cout << s.cvector[1] << std::endl;
    }

    return 0;
}

错误输出（跳过“必需”内容）：

error: ‘class arma::Col<double>::fixed<3ull>’ has no member named ‘serialize’; did you mean ‘set_size’? t.serialize(ar, file_version);

编辑： This帖子似乎与我的问题有关，遗憾的是它没有答案。

Answer 1

这里问题的真正症结在于你想为各种犰狳物体添加一个serialize()成员函数，但这看起来并不可能......除非是因为聪明在Armadillo中使用预处理器，它就是！

在Mat_bones.hpp和Col_bones.hpp的类定义中查看Mat和Col ...您会看到类似的内容：< / p>

public:

#ifdef ARMA_EXTRA_COL_PROTO
  #include ARMA_INCFILE_WRAP(ARMA_EXTRA_COL_PROTO)
#endif

当我发现这一点时，我非常高兴，因为现在我可以做一些像定义名为Mat_extra_bones.hpp的文件：

//! Add a serialization operator.
template<typename Archive>
void serialize(Archive& ar, const unsigned int version);

然后Mat_extra_meat.hpp：

// Add a serialization operator.
template<typename eT>
template<typename Archive>
void Mat<eT>::serialize(Archive& ar, const unsigned int /* version */)
{
  using boost::serialization::make_nvp;
  using boost::serialization::make_array;

  const uword old_n_elem = n_elem;

  // This is accurate from Armadillo 3.6.0 onwards.
  // We can't use BOOST_SERIALIZATION_NVP() because of the access::rw() call.
  ar & make_nvp("n_rows", access::rw(n_rows));
  ar & make_nvp("n_cols", access::rw(n_cols));
  ar & make_nvp("n_elem", access::rw(n_elem));
  ar & make_nvp("vec_state", access::rw(vec_state));

  // mem_state will always be 0 on load, so we don't need to save it.
  if (Archive::is_loading::value)
  {
    // Don't free if local memory is being used.
    if (mem_state == 0 && mem != NULL && old_n_elem > arma_config::mat_prealloc)
    {
      memory::release(access::rw(mem));
    }

    access::rw(mem_state) = 0;

    // We also need to allocate the memory we're using.
    init_cold();
  }

  ar & make_array(access::rwp(mem), n_elem);
}

然后，在您的程序中，您需要做的就是

#define ARMA_EXTRA_MAT_PROTO mat_extra_bones.hpp
#define ARMA_EXTRA_MAT_MEAT mat_extra_meat.hpp

并且serialize()函数将是Mat类的成员。您可以轻松地将此解决方案用于其他Armadillo类型。

实际上这正是mlpack库（http://www.mlpack.org/）所做的，所以如果你有兴趣，可以仔细看看我在那里实现的确切解决方案：

https://github.com/mlpack/mlpack/tree/master/src/mlpack/core/arma_extend

Answer 2

由于arm :: Col :: fixed不支持序列化本身，你可以在S类中编写它，或者编写一个包装它并将其序列化的类。我建议使用第二个选项，因为它可以让你在任何想要序列化的东西中使用arm :: Col :: fixed而不重复。

Answer 3

根据@UKMonkey的回答，我写了一个工作实例。实际上，对于这种情况，无需将serialize拆分为save和load。

#include <boost/mpi/environment.hpp>
#include <boost/mpi/communicator.hpp>
//#include <boost/serialization/split_free.hpp>
#include <iostream>
#include "armadillo"

namespace mpi = boost::mpi;

typedef arma::Col<double>::fixed<3> cvector;

//BOOST_SERIALIZATION_SPLIT_FREE(cvector)

namespace boost
{   
    namespace serialization
    {
        /*template<class Archive>
        void save(Archive& ar, const cvector& cv, unsigned int)
        {
            std::cout << "saving" << std::endl;
            ar& cv[0];
            ar& cv[1];
            ar& cv[2];
        }
        template<class Archive>
        void load(Archive& ar, cvector& cv, unsigned int)
        {
            std::cout << "loading" << std::endl;
            ar& cv[0];
            ar& cv[1];
            ar& cv[2];
        }*/      
        template<class Archive>
        inlide void serialize(Archive& ar, cvector& cv, unsigned int)
        {
            ar& cv[0];
            ar& cv[1];
            ar& cv[2];
        }
    }
} 

struct S
{
    int i;
    cvector c;

    friend class boost::serialization::access;

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

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

    S s;

    if (world.rank() == 0)
    {
        s.i = 3;
        s.c[0] = 2.;
        s.c[1] = 4.;
        world.send(1, 0, s);
    }
    else
    {
        world.recv(0, 0, s);
        std::cout << s.i << std::endl;
        std::cout << s.c[0] << std::endl;
        std::cout << s.c[1] << std::endl;
    }

    return 0;
}

如何序列化犰狳的矢量

3 个答案: