C ++

Question

我正在寻找是否有一种很好的“原生”方法来构造一个已知为类型PyObject*的对象。

这是我的代码：

C ++

void add_component(boost::python::object& type)
{
    auto constructed_type = type(); // doesn't construct anything!
}

的Python

o = GameObject()
o.add_component(CameraComponent)

我的代码完全正常执行整个函数，但永远不会触发CameraComponent的构造函数。

所以我的问题是，如果给定一个类型的PyObject*，我该如何构造该类型的实例？

非常感谢提前。

Answer 1

如果boost::python::object引用了一个类型，那么调用它将构造一个引用类型的对象：

boost::python::object type = /* Py_TYPE */;
boost::python::object object = type(); // isinstance(object, type) == True

由于Python中的几乎所有内容都是一个对象，因此接受来自Python的boost::python::object参数将允许任何类型的对象，即使是那些不是类型的对象。只要对象可调用（__call___），代码就会成功。

另一方面，如果要保证提供类型，那么一种解决方案是创建表示Python类型的C ++类型，接受它作为参数，并使用自定义转换器来构造C ++类型只有提供了Python类型。

以下type_object C ++类型表示一个Py_TYPE的Python对象。

/// @brief boost::python::object that refers to a type.
struct type_object: 
  public boost::python::object
{
  /// @brief If the object is a type, then refer to it.  Otherwise,
  ///        refer to the instance's type.
  explicit
  type_object(boost::python::object object):
    boost::python::object(object)
  {
    if (!PyType_Check(object.ptr()))
    {
      throw std::invalid_argument("type_object requires a Python type");
    }
  }
};

...

// Only accepts a Python type.
void add_component(type_object type) { ... }

以下自定义转换器只会在提供type_object PyObject*的情况下构建Py_TYPE个实例：

/// @brief Enable automatic conversions to type_object.
struct enable_type_object
{
  enable_type_object()
  {
    boost::python::converter::registry::push_back(
      &convertible,
      &construct,
      boost::python::type_id<type_object>());
  }

  static void* convertible(PyObject* object)
  {
    return PyType_Check(object) ? object : NULL;
  }

  static void construct(
    PyObject* object,
    boost::python::converter::rvalue_from_python_stage1_data* data)
  {
    // Obtain a handle to the memory block that the converter has allocated
    // for the C++ type.
    namespace python = boost::python;
    typedef python::converter::rvalue_from_python_storage<type_object>
                                                                 storage_type;
    void* storage = reinterpret_cast<storage_type*>(data)->storage.bytes;

    // Construct the type object within the storage.  Object is a borrowed 
    // reference, so create a handle indicting it is borrowed for proper
    // reference counting.
    python::handle<> handle(python::borrowed(object));
    new (storage) type_object(python::object(handle));

    // Set convertible to indicate success. 
    data->convertible = storage;
  }
};

...

BOOST_PYTHON_MODULE(...)
{
  enable_type_object(); // register type_object converter.
}

这是一个完整的示例demonstrating，它公开了一个需要Python类型的函数，然后构造了一个类型的实例：

#include <iostream>
#include <stdexcept> // std::invalid_argument
#include <boost/python.hpp>

/// @brief boost::python::object that refers to a type.
struct type_object: 
  public boost::python::object
{
  /// @brief If the object is a type, then refer to it.  Otherwise,
  ///        refer to the instance's type.
  explicit
  type_object(boost::python::object object):
    boost::python::object(object)
  {
    if (!PyType_Check(object.ptr()))
    {
      throw std::invalid_argument("type_object requires a Python type");
    }
  }
};

/// @brief Enable automatic conversions to type_object.
struct enable_type_object
{
  enable_type_object()
  {
    boost::python::converter::registry::push_back(
      &convertible,
      &construct,
      boost::python::type_id<type_object>());
  }

  static void* convertible(PyObject* object)
  {
    return PyType_Check(object) ? object : NULL;
  }

  static void construct(
    PyObject* object,
    boost::python::converter::rvalue_from_python_stage1_data* data)
  {
    // Obtain a handle to the memory block that the converter has allocated
    // for the C++ type.
    namespace python = boost::python;
    typedef python::converter::rvalue_from_python_storage<type_object>
                                                                 storage_type;
    void* storage = reinterpret_cast<storage_type*>(data)->storage.bytes;

    // Construct the type object within the storage.  Object is a borrowed 
    // reference, so create a handle indicting it is borrowed for proper
    // reference counting.
    python::handle<> handle(python::borrowed(object));
    new (storage) type_object(python::object(handle));

    // Set convertible to indicate success. 
    data->convertible = storage;
  }
};

// Mock API.
struct GameObject {};
struct CameraComponent
{
  CameraComponent()
  {
    std::cout << "CameraComponent()" << std::endl;
  }
};

boost::python::object add_component(GameObject& /* self */, type_object type)
{
  auto constructed_type = type();
  return constructed_type;
}

BOOST_PYTHON_MODULE(example)
{
  namespace python = boost::python;

  // Enable receiving type_object as arguments.
  enable_type_object();

  python::class_<GameObject>("GameObject")
    .def("add_component", &add_component);

  python::class_<CameraComponent>("CameraComponent");
}

交互式使用：

>>> import example
>>> game = example.GameObject()
>>> component = game.add_component(example.CameraComponent)
CameraComponent()
>>> assert(isinstance(component, example.CameraComponent))
>>> try:
...     game.add_component(component) # throws Boost.Python.ArgumentError
...     assert(False)
... except TypeError:
...     assert(True)
...

Answer 2

这实际上总是有效，但编译器正在以某种方式优化构造函数逻辑，所以我的断点永远不会被击中！

Python C API - 如何从PyObject构造对象

C ++

的Python

2 个答案: