Boost.Python

Header <boost/python/has_back_reference.hpp>

Class template has_back_reference

Class template has_back_reference synopsis

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.

Classes

Class template `has_back_reference`

A unary metafunction whose value is true iff its argument is a pointer_wrapper<>.

Class template `has_back_reference` synopsis

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_.

Example

C++ module definition

#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.

Python code

>>> 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)

Revised 18 July, 2004