Python C API - 如何从 PyObject 构造对象

Python C API - How to construct object from PyObject

本文关键字:PyObject 对象 API Python      更新时间:2023-10-16

我想知道是否有一种很好的"本机"方法来构造给定已知类型的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!
}

o = GameObject()
o.add_component(CameraComponent)

我的代码可以很好地执行整个函数,但构造函数永远不会被触发CameraComponent

所以我的问题是,给定已知为类型的PyObject*,我如何构造该类型的实例?

提前非常感谢。

如果boost::python::object引用类型,则调用它将使用引用的类型构造一个对象:

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对象,因此接受来自Python的参数作为boost::python::object将允许任何类型的对象,即使是那些不是类型的对象。 只要对象是可调用的(__call___),那么代码就会成功。


另一方面,如果要保证提供类型,则一种解决方案是创建一个表示 Python 类型的C++类型,接受它作为参数,并仅在提供 Python 类型时使用自定义转换器构造C++类型。

以下type_object C++类型表示一个 Python 对象,该对象是一个Py_TYPE

/// @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) { ... }

以下自定义转换器仅在提供Py_TYPE PyObject*时构造type_object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.
}

下面是一个完整的示例,演示如何公开一个需要 Python 类型的函数,然后构造该类型的实例:

#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");
}

交互式用法:

>>> 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)
...

这实际上总是有效的,但是编译器以某种方式优化了构造函数逻辑,所以我的断点从未被命中!