从元组解包参数 [英] Unpacking arguments from tuples
问题描述
So I'm trying to figure out how this works: C++11: I can go from multiple args to tuple, but can I go from tuple to multiple args?
我不明白的黑魔法这块代码片段：
The piece of black magic I do not understand is this code fragment:
f(std::get<N>(std::forward<Tuple>(t))...)
这是 f 里面的表达式，我不明白。
it's the expression inside f that I don't understand.
将 t 中的内容扩展为参数列表。但是有人可以解释这是怎么做的吗？当我看看 std :: get （ http://en.cppreference.com/w/cpp/utility/tuple/get ），我看不到 N 如何适合.. 。？据我所知，N是一个整数序列。
I understand that the expression somehow unpacks/expands what's inside t into a list of arguments. But could someone care to explain how this is done? When I look at the definition of std::get (http://en.cppreference.com/w/cpp/utility/tuple/get), I don't see how N fits in...? As far as I can tell, N is a sequence of integers.
根据我可以观察到的情况，我假设表达式 E 其中 X 是类型 X1 的序列。 X2 ，... Xn ，则表达式将扩展为 E < E X2。 ... E< Xn> 。这是怎么工作的？
Based on what I can observe, I'm assuming that expressions in the form E<X>... where X is the sequence of types X1. X2, ... Xn, the expression will be expanded as E<X1>, E<X2> ... E<Xn>. Is this how it works?
编辑：在这种情况下，N不是一个类型序列，而是整数。但我猜这个语言结构适用于类型和值。
Edit: In this case N is not a sequence of types, but integers. But I'm guessing this language construct applies to both types and values.
推荐答案
我认为@ Xeo的评论总结出来。从C ++ 11标准的14.5.3：
I think that @Xeo's comment summed it up well. From 14.5.3 of the C++11 standard:
包扩展包含一个模式 ，
实例化在列表中产生零个或多个
模式的实例化。
A pack expansion consists of a pattern and an ellipsis, the instantiation of which produces zero or more instantiations of the pattern in a list.
在你的情况下，当你完成递归模板实例化并结束部分专业化时，你有
In your case, by the time you finish with the recursive template instantiation and end up in the partial specialization, you have
f(std::get<N>(std::forward<Tuple>(t))...);
...其中 N 是参数包， c $ c> 2 和 3 ）。从上面的标准，模式在这里是
...where N is parameter pack of four ints (0, 1, 2, and 3). From the standardese above, the pattern here is
std::get<N>(std::forward<Tuple>(t))
。省略号到上述模式导致它被扩展成列表形式的四个实例化，即
The application of the ... ellipsis to the above pattern causes it to be expanded into four instantiations in list form, i.e.
f(std::get<0>(t), std::get<1>(t), std::get<2>(t), std::get<3>(t));
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