创建具有可变类型的元组 [英] Create a tuple with variatic type wrapped

问题描述

今天，我正在尝试在编译时创建一个特定的元组(至少对我来说).

Today I'm trying to create a tuple a little specific (for me at least) and at compile time.

我有一些基本的结构，可以说:

I have some basic struct, let say :

struct Foo1 { int data; };
struct Foo2 { int data; };
struct Foo3 { int data; };

另一个结构，但有一些模板内容:

And another struct but with some template stuff :

template < typename T,
           size_t Size >
struct Metadata {

  using type = T;

  std::bitset<Size>  bitset;
};

所以现在我想创建这种元组:

So now I want to create this kind of tuple :

constexpr std::tuple<Metadata<Foo1, 3>, Metadata<Foo2, 3>, Metadata<Foo3, 3>> test { {0}, {0}, {0}};

但是以一种自动方式，更像是:

But in an automatic way, more like:

template < typename ... Ts >
constexpr auto make_metadata() {

  return std::tuple<Metadata<Foo1, sizeof...(Ts)>, 
                    Metadata<Foo2, sizeof...(Ts)>,
                    Metadata<Foo3, sizeof...(Ts)>>{{0},{0},{0}};
}

最后一个代码远不是很好，所以我想拥有类似的功能，但是却是自动的.也许带有tuple_cat和fold表达式，但是我有点迷失了.因此，如果有人知道答案:)

Well the last code is far from good, so I'd like to have something like that but automatic. Maybe with tuple_cat and fold expression, but I'm a little lost. So if somebody know the answer :)

推荐答案

您可以使用...在单个表达式中表示多个内容.在这种情况下，您想立即展开Ts并立即展开:

You can use ... to mean multiple things in a single expression. In this case, you want to expand Ts both immediately and non-immediately:

template <class... Ts>
constexpr auto make_metadata()
{
    return std::make_tuple(Metadata<Ts, sizeof...(Ts)>{0}...);
}

此外，如果让您更清楚地以这种方式编写内容，则您不必将所有内容都写在一行上.

Also, you don't have to write everything on a single line if it makes it clearer for you to write it this way:

template <class... Ts>
constexpr auto make_metadata()
{
    constexpr size_t N = sizeof...(Ts);
    return std::make_tuple(Metadata<Ts, N>{0}...);
}

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