python/doc/reference/has_back_reference.qbk
2015-08-04 15:34:56 -04:00

116 lines
3.2 KiB
Plaintext

[section boost/python/has_back_reference.hpp]
[section Introduction]
<boost/python/has_back_reference.hpp> defines the predicate metafunction `has_back_reference<>`, which can be specialized by the user to indicate that a wrapped class instance holds a `PyObject*` corresponding to a Python object.
[endsect]
[section Class template `has_back_reference`]
A unary metafunction whose value is true iff its argument is a `pointer_wrapper<>`.
``
namespace boost { namespace python
{
template<class WrappedClass> class has_back_reference
{
typedef mpl::false_ type;
};
}}
``
A metafunction that is inspected by Boost.Python to determine how wrapped classes can be constructed.
`type::value` is an integral constant convertible to bool of unspecified type.
Specializations may substitute a true-valued integral constant wrapper for type iff for each invocation of `class_<WrappedClass>::def(init< type-sequence...>())` and the implicitly wrapped copy constructor (unless it is noncopyable), there exists a corresponding constructor `WrappedClass::WrappedClass(PyObject*, type-sequence...)`. If such a specialization exists, the WrappedClass constructors will be called with a "back reference" pointer to the corresponding Python object whenever they are invoked from Python. The easiest way to provide this nested type is to derive the specialization from `mpl::true_`.
[endsect]
[section Examples]
In C++:
``
#include <boost/python/class.hpp>
#include <boost/python/module.hpp>
#include <boost/python/has_back_reference.hpp>
#include <boost/python/handle.hpp>
#include <boost/shared_ptr.hpp>
using namespace boost::python;
using boost::shared_ptr;
struct X
{
X(PyObject* self) : m_self(self), m_x(0) {}
X(PyObject* self, int x) : m_self(self), m_x(x) {}
X(PyObject* self, X const& other) : m_self(self), m_x(other.m_x) {}
handle<> self() { return handle<>(borrowed(m_self)); }
int get() { return m_x; }
void set(int x) { m_x = x; }
PyObject* m_self;
int m_x;
};
// specialize has_back_reference for X
namespace boost { namespace python
{
template <>
struct has_back_reference<X>
: mpl::true_
{};
}}
struct Y
{
Y() : m_x(0) {}
Y(int x) : m_x(x) {}
int get() { return m_x; }
void set(int x) { m_x = x; }
int m_x;
};
shared_ptr<Y>
Y_self(shared_ptr<Y> self) { return self; }
BOOST_PYTHON_MODULE(back_references)
{
class_<X>("X")
.def(init<int>())
.def("self", &X::self)
.def("get", &X::get)
.def("set", &X::set)
;
class_<Y, shared_ptr<Y> >("Y")
.def(init<int>())
.def("get", &Y::get)
.def("set", &Y::set)
.def("self", Y_self)
;
}
``
The following Python session illustrates that x.self() returns the same Python object on which it is invoked, while y.self() must create a new Python object which refers to the same Y instance.
In Python:
``
>>> from back_references import *
>>> x = X(1)
>>> x2 = x.self()
>>> x2 is x
1
>>> (x.get(), x2.get())
(1, 1)
>>> x.set(10)
>>> (x.get(), x2.get())
(10, 10)
>>>
>>>
>>> y = Y(2)
>>> y2 = y.self()
>>> y2 is y
0
>>> (y.get(), y2.get())
(2, 2)
>>> y.set(20)
>>> (y.get(), y2.get())
(20, 20)
``
[endsect]
[endsect]