How would one call std::forward on all arguments in a variadic function?

Question

I was just writing a generic object factory and using the boost preprocessor meta-library to make a variadic template (using 2010 and it doesn't support them). My function uses rval references and std::forward to do perfect forwarding and it got me thinking...when C++0X comes out and I had a standard compiler I would do this with real variadic templates. How though, would I call std::forward on the arguments?

template <typename ...Params>
void f(Params... params) // how do I say these are rvalue reference?
{
    y(std::forward(...params)); //? - I doubt this would work.
}

Only way I can think of would require manual unpacking of ...params and I'm not quite there yet either. Is there a quicker syntax that would work?

Answer 1

You would do:

template <typename ...Params>
void f(Params&&... params)
{
    y(std::forward<Params>(params)...);
}

The ... is a pack expansion and pretty much says "take what's on the left, and for each template parameter, unpack it accordingly."

For example, f<int, float&> instantiated looks like:

void f<int, float&>(int&& params_0, float& params_1)
{
    y(std::forward<int>(params_0), std::forward<float&>(params_1));
}