使用boost-python将python变量设置为C++对象指针

Question

我想从C++设置一个Python变量,以便C++程序可以创建一个对象Game* game = new Game();,以便Python代码能够引用这个实例(和调用函数等).我怎样才能做到这一点？

我觉得我对Python或Boost-Python的工作方式有一些核心的误解.

该行在main_module.attr("game") = gametry catch语句中,并且错误(使用PyErr_Fetch)是"为C++类型找到No to_python(by-value)转换器:class Game".

例如

class_<Game>("Game")
        .def("add", &Game::add)
;

object main_module = import("__main__");
Game* game = new Game();
main_module.attr("game") = game; //This does not work

来自Python:

import testmodule

testmodule.game.foo(7)

Answer 1

在处理语言绑定时,通常必须在细节上迂腐.默认情况下,当C++对象超越语言边界时,Boost.Python将创建一个副本,因为这是防止悬空引用的最安全的操作过程.如果不应该创建副本,那么需要明确C++对象的所有权:

• 要在保持C++所有权的同时将对C++对象的引用传递给Python,请使用boost::python::ptr()或boost::ref().C++代码应该保证C++对象的生命周期至少和Python对象一样长.使用时ptr(),如果指针为null,则生成的Python对象将为None.
• 要将C++对象的所有权转移到Python,可以应用ResultConverterGenerator,允许将所有权转移到Python.一旦Python对象的生命周期结束,C++代码不应尝试访问指针.manage_new_object
• 对于共享所有权,需要使用支持共享语义的智能指针的HeldType公开该类,例如boost::shared_ptr.

一旦创建了Python对象,就需要将其插入到Python命名空间中以便通常可访问:

• 在模块定义中,用于boost::python::scope获取当前范围的句柄.例如,以下内容将插入x到example模块中:

  BOOST_PYTHON_MODULE(example)
  {
    boost::python::scope().attr("x") = ...; // example.x
  }
  

• 要插入__main__模块,可以导入__main__.例如,以下内容将插入x到__main__模块中:

  boost::python::import("__main__").attr("x") = ...;
  

下面是一个演示如何从C++直接构造Python对象,将C++对象的所有权转移到Python,以及构造引用C++对象的Python对象的示例:

#include <iostream>
#include <boost/python.hpp>

// Mockup model.
struct spam
{
  spam(int id)
    : id_(id)
  {
    std::cout << "spam(" << id_ << "): "  << this << std::endl;
  }

  ~spam()
  {
    std::cout << "~spam(" << id_ << "): " << this << std::endl;
  }

  // Explicitly disable copying.
  spam(const spam&) = delete;
  spam& operator=(const spam&) = delete;

  int id_;
};

/// @brief Transfer ownership to a Python object.  If the transfer fails,
///        then object will be destroyed and an exception is thrown.
template <typename T>
boost::python::object transfer_to_python(T* t)
{
  // Transfer ownership to a smart pointer, allowing for proper cleanup
  // incase Boost.Python throws.
  std::unique_ptr<T> ptr(t);

  // Use the manage_new_object generator to transfer ownership to Python.
  namespace python = boost::python;
  typename python::manage_new_object::apply<T*>::type converter;

  // Transfer ownership to the Python handler and release ownership
  // from C++.
  python::handle<> handle(converter(*ptr));
  ptr.release();

  return python::object(handle);
}

namespace {
spam* global_spam;
} // namespace

BOOST_PYTHON_MODULE(example)
{
  namespace python = boost::python;
  // Expose spam.
  auto py_spam_type = python::class_<spam, boost::noncopyable>(
      "Spam", python::init<int>())
    .def_readonly("id", &spam::id_)
    ;

  // Directly create an instance of Python Spam and insert it into this
  // module's namespace.
  python::scope().attr("spam1") = py_spam_type(1);

  // Construct of an instance of Python Spam from C++ spam, transfering
  // ownership to Python.  The Python Spam instance will be inserted into
  // this module's namespace.
  python::scope().attr("spam2") = transfer_to_python(new spam(2));

  // Construct an instance of Python Spam from C++, but retain ownership of
  // spam in C++.  The Python Spam instance will be inserted into the
  // __main__ scope.
  global_spam = new spam(3);
  python::import("__main__").attr("spam3") = python::ptr(global_spam);
}

互动用法:

>>> import example
spam(1): 0x1884d40
spam(2): 0x1831750
spam(3): 0x183bd00
>>> assert(1 == example.spam1.id)
>>> assert(2 == example.spam2.id)
>>> assert(3 == spam3.id)
~spam(1): 0x1884d40
~spam(2): 0x1831750

在示例用法中,请注意Python spam(3)在退出时如何不销毁,因为它未被授予底层对象的所有权.