C++ 如何对可变参数函数中的所有参数调用 std::forward？
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How would one call std::forward on all arguments in a variadic function?
提问by Edward Strange
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::forwardto 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::forwardon the arguments?
我只是在编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 并且它不支持它们）。我的函数使用 rval 引用并std::forward进行完美转发，这让我开始思考……当 C++0X 出现并且我有一个标准编译器时，我会用真正的可变参数模板来做到这一点。但是，我会如何呼吁std::forward争论？
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?
我能想到的唯一方法是手动解包 ...params ，而且我还没有到那里。有没有更快的语法可以工作？
回答by GManNickG
You would do:
你会这样做：
template <typename ...Params>
void f(Params&&... params)
{
y(std::forward<Params>(params)...);
}
The ...pretty much says "take what's on the left, and for each template parameter, unpack it accordingly."
在...相当多说，“采取什么样的左边，并为每个模板参数，因此解压。”
