将 C++ 创建的对象追加到 python 列表并使其由 python 管理

Wer*_*ner 2 boost memory-management list boost-python

好吧,我已经检查了一段时间,找不到答案。

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我想附加一个暴露给 python 的对象,比如 Foo:

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struct Foo {\n  Foo(){ std::cout << "Creating a Foo object" << std::endl;}\n  virtual ~Foo(){ std::cout << "Destroying a Foo object" << std::endl;}\n};\n
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我使用 Foo 继承的对象,有时我想将它们附加到 python 列表中。为此,我创建了一个 FooWrapper,它继承自 Foo 并使用复制构造函数

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struct FooWrapper : public Foo {\n  FooWrapper(const Foo& foo):Foo(foo){ std::cout << "Creating a copy from foo using FooWrapper" << std::endl;}  \n  virtual ~FooWrapper(){ std::cout << "Destroying a FooWrapper object" << std::endl;}\n};\n
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这是暴露给Python的:

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BOOST_PYTHON_MODULE(foo_module)\n{\n    using namespace py = boost::python;\n    py::class_<FooWrapper>("FooWrapper", py::no_init)\xe2\x80\xa6\n}\n
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我有一个方法将最终的 Foo 对象作为 FooWrapper 附加到 python 列表中,例如:

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void appendFoosToList(py::list &list)\n{\n  for ( const Foo* foo : m_foos )\n  {\n    list.append( FooWrapper( *foo )  );\n  }\n}                                                                                                                                 \n
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我怎样才能做到这一点,而不是创建一个临时对象然后复制到列表,而是将该对象附加到列表中,而不必复制临时对象?

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我读过很多文档(boost_faqboost_python_wiki),很多次我遇到这个运行时错误:

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类型错误: 找不到 C++ 类型的 to_python (按值) 转换器:\n

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BPL 无法从 Python 对象获取 C++ 值。

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例如,当调用 extract(.attr(" len ")()) \n 来获取对象长度时,您省略了“()”。

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并且没能找到解决方案。

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我找不到关于此的明确文档,所以我来到这里作为最后的手段。

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Tan*_*ury 5

简而言之,在空闲存储上分配包装器并使用结果manage_new_object转换将所有权转移到 Python 对象。这将导致 Boost.Python 在构造 Python 对象时复制指针,而不是复制被指针。

C++ 对象嵌入到 Python 对象中。这是HeldType通过 公开类时提供的class_,默认为公开的 C++ 类型。通常,在公开函数时,可以使用CallPolicy实例来扩充返回的 C++ 类型并将其嵌入到 Python 对象中。特别是,将 CallPolicy 实例return_value_policyResultConverterGenerator一起使用允许嵌入类型成为指针,并且 Python 对象将管理所有权。manage_new_object

以下函数可用于将对象的所有权转移到 Python,而无需复制被指向对象:

/// @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);
}
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用法示例:

void appendFoosToList(boost::python::list& list)
{
  for (const Foo* foo : m_foos)
  {
    list.append(transfer_to_python(new FooWrapper(*foo)));
  }
}
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这是演示此方法的完整示例:

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

// Mocks...
class spam
{
public:
  spam() { std::cout << "spam(): " << this << std::endl; }
  spam(const spam&)
  {
    std::cout << "spam(const spam&): " << this << std::endl;
  }
  ~spam() { std::cout << "~spam(): " << this << std::endl; }
};

/// @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);
}

void append_to_list(boost::python::list& list)
{
  list.append(transfer_to_python(new spam()));
}

BOOST_PYTHON_MODULE(example)
{
  namespace python = boost::python;
  python::class_<spam>("Spam", python::no_init);
  python::def("append_to_list", &append_to_list);
}
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互动使用:

>>> import example
>>> spams = []
>>> example.append_to_list(spams)
spam(): 0x25cbd90
>>> assert(type(spams[0]) is example.Spam)
>>> del spams
~spam(): 0x25cbd90
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