Boost.python with universal reference / perfect forwarding?

Question

I have been experimenting with C++11 universal references and perfect forwarding in some classes that also need to be accessed from Python using Boost.Python. I have code that works, but it requires some ugly template specialization outside of the class. Has anybody solved this in a more elegant way? Does anybody have any suggestions on improving the following code?

#include <boost/python.hpp>
#include <string>

using namespace boost::python;

struct A {
    A() : _a("initial") {}

    template <typename T>
    void set_a(T&& a) { _a = std::forward<T>(a); }

    const std::string& get_a() const { return _a; }

private:
    std::string _a;
};

// How can the following template member function specialization be avoided?
template <>
void A::set_a(const std::string& a) { _a = a; }

BOOST_PYTHON_MODULE(example)
{
    class_<A>("A")
       .add_property("a", make_function( &A::get_a, return_value_policy<copy_const_reference>()),
                     &A::set_a<const std::string&>) // Can this be defined differently?
    ;
}

Answer 1

I experimented some today and found the answer is actually pretty simple. Just use make_function() again in the setr part of add_property:

Here is the simplified code:

#include <boost/python.hpp>
#include <string>

using namespace boost::python;

struct A {
    A() : _a("initial") {}

    template <typename T>
    void set_a(T&& a) { _a = std::forward<T>(a); }

    const std::string& get_a() const { return _a; }

private:
    std::string _a;
};

BOOST_PYTHON_MODULE(example)
{
    class_<A>("A")
       .add_property("value",
                     make_function( &A::get_a, return_value_policy<copy_const_reference>()),
                     make_function( &A::set_a<const std::string&> )
    );
}