复制boost.python对象

我有一些boost python类,我在python中实例化.我想复制它们.所以,如果我有

p = Bernoulli(0.5)

我想要做

q = Bernoulli(p)

但是,如果我不知道p的类型怎么办？我试着这样做：

q = copy.deepcopy(p)

但是python说它无法腌制p.

我唯一的解决方案是将一个clone()函数添加到Bernoulli的接口中吗？或者我可以以某种方式自动生成该方法吗？ copy.deepcopy可以与Boost.python对象一起使用吗？

解决方法:

从http://mail.python.org/pipermail/cplusplus-sig/2009-May/014505.html起

#define PYTHON_ERROR(TYPE, REASON) \
{ \
    PyErr_SetString(TYPE, REASON); \
    throw bp::error_already_set(); \
}

template<class T>
inline PyObject * managingPyObject(T *p)
{
    return typename bp::manage_new_object::apply<T *>::type()(p);
}

template<class Copyable>
bp::object
generic__copy__(bp::object copyable)
{
    Copyable *newCopyable(new Copyable(bp::extract<const Copyable
&>(copyable)));
    bp::object
result(bp::detail::new_reference(managingPyObject(newCopyable)));

    bp::extract<bp::dict>(result.attr("__dict__"))().update(
        copyable.attr("__dict__"));

    return result;
}

template<class Copyable>
bp::object
generic__deepcopy__(bp::object copyable, bp::dict memo)
{
    bp::object copyMod = bp::import("copy");
    bp::object deepcopy = copyMod.attr("deepcopy");

    Copyable *newCopyable(new Copyable(bp::extract<const Copyable
&>(copyable)));
    bp::object
result(bp::detail::new_reference(managingPyObject(newCopyable)));

    // HACK: copyableId shall be the same as the result of id(copyable)
in Python -
    // please tell me that there is a better way! (and which ;-p)
    int copyableId = (int)(copyable.ptr());
    memo[copyableId] = result;

    bp::extract<bp::dict>(result.attr("__dict__"))().update(
        deepcopy(bp::extract<bp::dict>(copyable.attr("__dict__"))(),
memo));

    return result;
}

要使用它：

class_<foo>(foo)
   .def("__copy__", &generic__copy__< foo >)
   .def("__deepcopy__", &generic__deepcopy__< foo >)
   .def(init< const foo & >())

标签：boost-python,python
来源： https://codeday.me/bug/20190827/1737753.html