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Library/boost/libs/python/doc/reference/has_back_reference.qbk
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PatoFlamejanteTV
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19 дек 2024, 19:11
19 дек 2024, 19:11
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[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]