C++ weak functor solution

Question

This is specific to void() functors, but that is fine with me...

struct Foo
{
   void Bar(int x)
   {
      std::cout << x << std::endl;
   }
};

struct VoidBind
{
    typedef void result_type;

    template<typename T> void operator()(const std::weak_ptr<T>& obj, std::function<void()>& func)
    {
        std::shared_ptr<T> shared_obj = obj.lock();

        if (shared_obj)
        {
            func();
        }
    }

    template<typename T> static std::function<void()> bind(const std::shared_ptr<T>& obj, const std::function<void()>& func)
    {
        return std::bind(VoidBind(), std::weak_ptr<T>(obj), func);
    }
};

#define BIND(F, O, A...)  VoidBind::bind(O, std::function<void()>(std::bind(F, O.get(), ##A)))

This code is then invoked as...

auto obj = std::make_shared<Foo>();
auto func = BIND(&Foo::Bar, obj, 99);  // match std::bind style
func();
obj.reset();  // destroy object
func();       // will not do anything

My question is whether there is some way to avoid the BIND macro?

Answer 1

Variadic templates can be used here:

template<class F, class O, class... Args>
auto bind(F f, O&& o, Args&&... args)
    -> decltype(VoidBind::bind(o, std::function<void()>(std::bind(f, O.get(), args...))))
{
    return VoidBind::bind(O, std::function<void()>(std::bind(F, O.get(), args...)));
}

It gets better with C++14 automatic return-type deduction where you don't even need to specify the return value:

template<class F, class O, class... Args>
auto bind(F f, O&& o, Args&&... args)
{
    return VoidBind::bind(o, std::function<void()>(std::bind(f, O.get(), args...)));
}