使用pybind11共享MPI通信器

Question

假设我在MPI通信器周围创建了一个包装器：

class Communicator {
   public:
      Communicator() : comm(MPI_COMM_WORLD) {}

      Communicator(int const color, int const key) {
        MPI_Comm_split(MPI_COMM_WORLD, color, key, &comm);
      }

      Communicator(MPI_Comm comm) : comm(comm) {}

      MPI_Comm GetComm() const { return comm; }
    private:
      MPI_Comm comm;
}; 

我想使用pybind11围绕这个对象创建一个python包装器，如下所示：

void CommunicatorWrapper(pybind11::module &m) {
   py::class_<Communicator, std::shared_ptr<Communicator> > commWrap(m, "Communicator");

   commWrap.def(py::init( []() { return new Communicator(); } ));
   commWrap.def(py::init( [](int const color, int const key) { return new Communicator(color, key); } ));
   commWrap.def(py::init( [](MPI_Comm comm) { return new Communicator(comm); } ));
   commWrap.def("GetComm", &Communicator::GetComm);
}

但是，我希望python看到的MPI_Comm类型是mpi4py.MPI.Comm。这可能吗？如果是，怎么做？你知道吗

上述（原始）实现会导致以下行为：

comm = Communicator(MPI.COMM_WORLD)

错误：

TypeError: __init__(): incompatible constructor arguments. The following argument types are supported:
1. Communicator()
2. Communicator(arg0: int, arg1: int)
3. Communicator(arg0: int)

以及

comm = Communicator()
print(comm.GetComm())

打印-2080374784。考虑到MPI_Comm是什么，这种行为是有意义的，但显然不是我需要的功能。你知道吗

Answer 1

我把包装纸改成

#include <mpi4py/mpi4py.h>

pybind11::handle CallGetComm(Communicator *comm) {
    const int rc = import_mpi4py();
    return pybind11::handle(PyMPIComm_New(comm->GetComm()));;
}

void CommunicatorWrapper(pybind11::module &m) {
   py::class_<Communicator, std::shared_ptr<Communicator> > commWrap(m, "Communicator");

   commWrap.def(py::init( []() { return new Communicator(); } ));
   commWrap.def(py::init( [](int const color, int const key) { return new Communicator(color, key); } ));
   commWrap.def(py::init( [](pybind11::handle const& comm) {
     const int rc = import_mpi4py();
     assert(rc==0);
     return new Communicator(*PyMPIComm_Get(comm.ptr()));
    } ));
   commWrap.def("GetComm", &CallGetComm);
}