How would one call std::forward on all arguments in a variadic function?
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by Noah Roberts
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Published on 2010-05-12T17:28:11Z
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2010/05/12
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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?
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