我正在尝试将C ++容器暴露给Python。 我有:
class Container {
std::auto_ptr<Iterator> __iter__();
};
class Iterator {
Container & parent;
Item __next__();
};
class Item {
Container & parent;
};
Item
类在内部引用Container中存在的数据。返回Iterator
实例的Item
不一定存在,Item
可用。
c = Container()
for i in c:
store = i
print store
在上面的代码中,我希望获得Container
,Iterator
和少量Item
个实例。
当它达到print
语句时,我希望Iterator
已经被销毁,但Container
显然仍然存在store
个实例。
现在问题来了。我不知道用CallPolicy
来实现这种效果:
定义:
class_<Container>("Container", ...)
.def("__iter__", &Container::__iter__, return_interal_reference<>() )
;
class_<Iterator>("Iterator", ...)
.def("next", &Iterator::__next__, what_call_policy_here? )
;
class_<Item>("Item", ...)
.def("__str__", ... )
;
我应该使用什么代替what_call_policy_here
?
答案 0 :(得分:2)
好的,经过长时间的挖掘后,我想我想出了一个对暴露类型透明的解决方案。
基本上解决方案是创建CallPolicy
,它将自动存储对返回的子对象内的父对象(即Container
)的引用(即{ {1}})作为它的属性(我使用了私有名称,但Python在这方面非常自由)。
然后自动将其复制到所有兄弟对象(兄弟是父的另一个子,但是一个是通过调用另一个子的方法创建的,因此不是直接来自 parent )。
这需要摆弄Iterator
。我必须创建两个自定义的:
CallPolicy
现在用法:
// This policy is used for methods returning items that require object to be
// kept as long as return thing is alive.
// It stores reference in attribute named Property_::name
template <typename Property_, class BasePolicy_ = boost::python::default_call_policies>
struct store_parent_reference: public BasePolicy_
{
template <class ArgumentPackage>
static PyObject* postcall(ArgumentPackage const& args_, PyObject* result)
{
result = BasePolicy_::postcall(args_, result);
PyObject* parent = detail::get_prev< std::size_t(1) >::execute(args_, result);
PyObject* child = result;
if( PyObject_SetAttrString( child, Property_::name, parent ) == -1 )
{
std::ostringstream err;
err << "store_parent_reference::postcall could not set attribute `" << Property_::name
<< "` on newly allocated object `"
<< extract<std::string>( object( handle<>(borrowed(child))).attr("__str__")() )()
<< "`";
throw std::runtime_error(err.str());
}
return result;
}
};
// This policy is used for methods returning "sibling" in the meaning both the returned object
// and one that has this method called on require "parent" object to be alive.
//
// It copies reference to "parent" to attribute named ChildProperty_::name
// from "original" object's attribute named SiblingProperty_::name
template <typename ChildProperty_, typename SiblingProperty_ = ChildProperty_, class BasePolicy_ = boost::python::default_call_policies>
struct copy_parent_from_sibling: public BasePolicy_
{
template <class ArgumentPackage>
static PyObject* postcall(ArgumentPackage const& args_, PyObject* result)
{
result = BasePolicy_::postcall(args_, result);
PyObject* sibling = detail::get_prev< std::size_t(1) >::execute(args_, result);
PyObject* new_child = result;
PyObject* parent = PyObject_GetAttrString( sibling, SiblingProperty_::name );
if( parent == NULL )
{
std::ostringstream err;
err << "copy_parent_from_sibling::postcall could not get attribute `"
<< SiblingProperty_::name
<< "` from sibling `"
<< extract<std::string>( object( handle<>(borrowed(sibling))).attr("__str__")() )()
<< "` to set up attribute `"
<< ChildProperty_::name
<< "` of returned object which is `"
<< extract<std::string>( object( handle<>(borrowed(new_child))).attr("__str__")() )()
<< "`";
throw std::runtime_error(err.str());
}
if( PyObject_SetAttrString( new_child, ChildProperty_::name, parent ) == -1 )
{
std::ostringstream err;
err << "copy_parent_from_sibling::postcall could not set attribute `"
<< ChildProperty_::name
<< "` on returned object which is `"
<< extract<std::string>( object( handle<>(borrowed(new_child))).attr("__str__")() )()
<< "`";
throw std::runtime_error(err.str());
}
Py_DECREF(parent);
return result;
}
};
注意:很难为struct ContainerProperty {
static const char * const name;
};
const char * const ContainerProperty::name = "__container"
class_<Container>("Container", ...)
.def("__iter__", &Container::__iter__, store_parent_reference< ContainerProperty >() )
;
class_<Iterator>("Iterator", ...)
.def("next", &Iterator::__next__, copy_parent_from_sibling< ContainerProperty >() )
;
class_<Item>("Item", ...)
;
内容提供完整的最小工作样本,所以我可能错过了上面的一些细节,但解决方案似乎对我很好(我是跟踪析构函数调用以检查)。
这也不是唯一的解决方案。请注意,boost::python
有点类似于store_parent_reference
,区别在于它明确需要存储数据的位置。这只是因为return_internal_reference
需要从某处复制它。
这种方法的主要好处是它不需要原始类来了解Python的东西。