如何对可变函数中的所有参数调用std :: forward？ [英] How would one call std::forward on all arguments in a variadic function?

问题描述

我只是写一个通用对象工厂，并使用boost预处理器元库来制作一个可变参数模板（使用2010，它不支持它们）。我的函数使用rval引用和std :: forward做完美的转发，它让我想...当C ++ 0X出来，我有一个标准的编译器，我会做真正的可变模板。如何，我可以在参数上调用std :: forward？

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）//我如何说这些是rvalue引用？
{
y（std :: forward（... params））; //？ - 我怀疑这将工作。
}

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

只有我想到的方式才需要手动解包。 ..params和我还不完全在那里。是否有更快的语法可以工作？

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?

推荐答案

你会这样做：

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

... 几乎说拿左边的，并为每个模板参数，相应地解压缩。

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

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