如何使用boost :: python将std :: pair暴露给python？

Question

如何std::pair使用boost::python？当我曝光时,vector<string>我只需写:

class_<std::vector<std::string> >("StringVec")
    .def(vector_indexing_suite<std::vector<std::string> >())
;

但我不知道如何处理std :: pair.

Answer 1

我找到了解决方案.暴露的最简单的例子std::pair是:

class_<std::pair<int, int> >("IntPair")
    .def_readwrite("first", &std::pair<int, int>::first)
    .def_readwrite("second", &std::pair<int, int>::second);

Answer 2

我使用以下代码公开std::pair<>为Python元组:

#include <boost/python.hpp>
namespace py = boost::python

template<typename T1, typename T2>
struct PairToPythonConverter {
    static PyObject* convert(const std::pair<T1, T2>& pair)
    {
        return py::incref(py::make_tuple(pair.first, pair.second).ptr());
    }
};

template<typename T1, typename T2>
struct PythonToPairConverter {
    PythonToPairConverter()
    {
        py::converter::registry::push_back(&convertible, &construct, py::type_id<std::pair<T1, T2> >());
    }
    static void* convertible(PyObject* obj)
    {
        if (!PyTuple_CheckExact(obj)) return 0;
        if (PyTuple_Size(obj) != 2) return 0;
        return obj;
    }
    static void construct(PyObject* obj, py::converter::rvalue_from_python_stage1_data* data)
    {
        py::tuple tuple(py::borrowed(obj));
        void* storage = ((py::converter::rvalue_from_python_storage<std::pair<T1, T2> >*) data)->storage.bytes;
        new (storage) std::pair<T1, T2>(py::extract<T1>(tuple[0]), py::extract<T2>(tuple[1]));
        data->convertible = storage;
    }
};

template<typename T1, typename T2>
struct py_pair {
    py::to_python_converter<std::pair<T1, T2>, PairToPythonConverter<T1, T2> > toPy;
    PythonToPairConverter<T1, T2> fromPy;
};

在我的主要内容中BOOST_PYTHON_MODULE(),我可以使用eg py_pair<int, int>();来暴露一对整数.