Python C API-如何从PyObject构造对象

我正在寻找一种已知的PyObject *是否存在一种很好的“本机”方式来构建对象.

这是我目前的代码：

C

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

Python

o = GameObject()
o.add_component(CameraComponent)

我的代码完美地执行了整个功能,但从未为CameraComponent触发构造函数.

所以我的问题是,给定一个已知为PyObject *的类型,我该如何构造该类型的实例？

提前谢谢了.

解决方法:

如果boost :: python :: object引用了一个类型,则对其进行调用将构造一个具有所引用类型的对象：

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

由于Python中几乎所有事物都是对象,因此将Python的参数接受为boost :: python :: object会允许任何类型的对象,即使不是类型的对象也是如此.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,将其接受为参数,并仅在以下情况下使用自定义转换器构造C类型：提供了Python类型.

以下type_object C类型表示一个Python对象,即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) { ... }

以下自定义转换器将仅在为type_object实例提供PyObject *作为Py_TYPE的情况下构造它：

/// @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");
}

互动用法：

>>> 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)
...

标签：boost-python,boost,python,c-4
来源： https://codeday.me/bug/20191119/2033886.html