再次迭代一个元组... [英] Iterating on a tuple... again

问题描述

过去一段时间我一直在做C ++，但我不熟悉模板。

It's been a while that I've been doing C++ but I'm not familiar with templates.

最近，我试图写一个类， code> std :: vector< std :: tuple< Types ...>> 。这个类必须有成员函数，我真的需要能够在元组上迭代。事实上，如果我能够打印一个元组的每个元素（按顺序），我就能够做我需要的一切。

Recently, I tried to write a class that wrap a std::vector<std::tuple<Types...>>. This class must have member functions, and I really need to be able to iterate over the tuple. In fact, if I am able to print every element of a tuple (in the order), I would be able to do everything I need.

我发现一个解决方案使用一个演员，但我不是真的有信心，因为它是基于一个演员，我不喜欢（加上，当我尝试使用 static_cast ，它不再编译）。

I found a solution using a cast, but I'm not really confident with it since it is based on a cast that I don't really like (plus, when I try to use static_cast, it doesn't compile anymore).

我的问题是，以下代码正确，可移植，它是一个黑客，我应该找到另一种方法来做这个比使用这个演员？此外，这个转换可能是一个运行时转换的权利？有没有办法做我想要的没有这个？

My question is, is the following code correct, portable, is it a hack and should I find another way to do this than to use this cast ? Also, this cast is probably a runtime-cast right ? Is there a way to do what I want without this ?

std::ostream& operator<<(std::ostream& out, std::tuple<> const& tuple)
{
    return out; // Nothing to do here
}

template<typename First, typename... Types>
std::ostream& operator<<(std::ostream& out, std::tuple<First, Types...> const& tuple)
{
    out << std::get<0>(tuple) << " ";

    // The cast that I don't like
    return out << (std::tuple<Types...>&) tuple;
}

int main()
{
    auto tuple = std::make_tuple(1, 2.3, "Hello");
    std::cout << tuple << std::endl;
    return 0;
}

预先感谢您的答复。

推荐答案

使用 <$ c

Use std::index_sequence_for for fun and profit.

template <typename TupleLike, size_t ... Inds>
std::ostream& PrintHelper(std::ostream& out, TupleLike const& tuple, std::index_sequence<Inds...>)
{
  int unused[] = {0, (void(out << std::get<Inds>(tuple) << " "), 0)...};
  (void)unused;
  return out;
}

template<typename... Types>
std::ostream& operator<<(std::ostream& out, std::tuple<Types...> const& tuple)
{
  return PrintHelper(out, tuple, std::index_sequence_for<Types...>());
}

编辑：现场演示。感谢@dyp。这使用了此答案的扩展技巧。

EDIT : Live Demo. Thanks to @dyp. This uses an expansion trick from this answer.

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