是否可以对参数包进行typedef？ [英] Is it possible to typedef a parameter pack?

问题描述

是否可以对参数包进行typedef？例如

Is it possible to typedef a parameter pack? For example

template<class T, class... Args>
struct A
{
    typedef T Type; // We typedef it, then its derived class can use it.
                    // How about for parameter packs?

    // Option 1:
    typedef Args Arguments;

    // Option 2:
    using Arguments = Args;

    // Option 3: I can put in a tuple, but how can I untuple it to a pack
    typedef tuple<Args...> Tuple;
};

我想使用上述技术来实现以下

I want to using the above technique to implement the following

template<int... VALUES>
struct IntegralSequence
{
    enum { SIZE = sizeof...(VALUES) };

    template <unsigned I>
    struct At
    {
        enum { VALUE = typename tuple_element<I, 
                       tuple<integral_constant<int, VALUES>...>>::type::value
             };
    };
};

template<unsigned N>
struct AscendingSequence
{
    typedef IntegralSequence<AscendingSequence<N-1>::VALUES..., N> Type;
    using VALUES = Type::VALUES; // if it works
};

template<>
struct AscendingSequence<1>
{
    typedef IntegralSequence<0> Type;
    using VALUES = Type::VALUES; // if it works
};

推荐答案

您可以将它们打包在元组，或在任意的空类模板中（我更喜欢将其称为 pack ）：

You can pack them in a tuple, or in a arbitrary empty class template (I prefer to call it pack):

template<typename... Args>
struct pack { };

template<class T, class... Args>
struct A
{
    using args = pack<Args...>;
};

如果给了 A 例如在函数模板中，并且您想推断出 Args ... ，您可以这样操作：

If you are then given A e.g. in function template and you want to deduce Args..., you can do it like this:

template<typename... Args, typename A>
void f(pack<Args...>, A a) { /* use Args... here */ }

template<typename A>
void f(A a) { f(typename A::args(), a); }

pack 在情况下很方便像那样。否则，您需要其他一些方法来传递 args 而不实际传递包含数据的 tuple （例如，将其包装到

pack being empty is convenient in situations like that. Otherwise you'd need some other means to pass args without actually passing a tuple that contains data (e.g. wrapping it into yet another empty struct).

或者在类模板专业化中：

Or, in a class template specialization:

template<typename T, typename = typename T::args>
struct B_impl;

template<typename T, typename... Args>
struct B_impl <T, pack<Args...> >
{
    // use Args... here
};

template<typename T>
using B = B_impl<T>;

我想这是@dyp提到的演绎和部分专业化的选择。

I guess these are the options of deduction and partial specialization that @dyp mentioned.

编辑这是对已编辑问题的答复。好的，这显然是XY问题。如果只需要 IntegralSequence ，则可以在C ++ 14中使用 std :: make_integer_sequence 或检查我对另一个问题的回答。

EDIT This is in response to the edited question. Ok, this is clearly an XY problem. If IntegralSequence is all you need, you can use std::make_integer_sequence in C++14 or check my answer to another question just a few minutes ago for an efficient implementation.

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