是否可以在不扩展的情况下“存储”模板参数包？

当我偶然发现此问题时，我正在尝试使用C ++ 0x可变参数模板：

template < typename ...Args >
struct identities
{
    typedef Args type; //compile error: "parameter packs not expanded with '...'
};

//The following code just shows an example of potential use, but has no relation
//with what I am actually trying to achieve.
template < typename T >
struct convert_in_tuple
{
    typedef std::tuple< typename T::type... > type;
};

typedef convert_in_tuple< identities< int, float > >::type int_float_tuple;

当我尝试对模板参数包进行typedef时，GCC 4.5.0给我一个错误。

基本上，我想将参数包“存储”在typedef中，而不将其解包。可能吗？如果不是，是否有某些原因不允许这样做？

比Ben的方法更通用的另一种方法如下：

#include <tuple>

template <typename... Args>
struct variadic_typedef
{
    // this single type represents a collection of types,
    // as the template arguments it took to define it
};

template <typename... Args>
struct convert_in_tuple
{
    // base case, nothing special,
    // just use the arguments directly
    // however they need to be used
    typedef std::tuple<Args...> type;
};

template <typename... Args>
struct convert_in_tuple<variadic_typedef<Args...>>
{
    // expand the variadic_typedef back into
    // its arguments, via specialization
    // (doesn't rely on functionality to be provided
    // by the variadic_typedef struct itself, generic)
    typedef typename convert_in_tuple<Args...>::type type;
};

typedef variadic_typedef<int, float> myTypes;
typedef convert_in_tuple<myTypes>::type int_float_tuple;

int main()
{}

我认为不允许这样做的原因是它会很杂乱，您可以解决它。您需要使用依赖关系反转，并使将参数包存储到工厂模板的结构能够将该参数包应用于另一个模板。

类似于以下内容：

template < typename ...Args >
struct identities
{
    template < template<typename ...> class T >
    struct apply
    {
        typedef T<Args...> type;
    };
};

template < template<template<typename ...> class> class T >
struct convert_in_tuple
{
    typedef typename T<std::tuple>::type type;
};

typedef convert_in_tuple< identities< int, float >::apply >::type int_float_tuple;

我发现Ben Voigt的想法对我自己的工作非常有用。我已经对其进行了少许修改，以使其不仅适用于元组。对这里的读者来说，这可能是一个明显的修改，但可能值得展示：

template <template <class ... Args> class T, class ... Args>
struct TypeWithList
{
  typedef T<Args...> type;
};

template <template <class ... Args> class T, class ... Args>
struct TypeWithList<T, VariadicTypedef<Args...>>
{
  typedef typename TypeWithList<T, Args...>::type type;
};

名称TypeWithList源自以下事实：现在已使用先前的列表实例化了该类型。

这是GManNickG巧妙的部分专业化技巧的变体。没有委托，并且通过要求使用variadic_typedef结构，可以提高类型安全性。

#include <tuple>

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

template<typename... Args>
struct convert_in_tuple {
    //Leaving this empty will cause the compiler
    //to complain if you try to access a "type" member.
    //You may also be able to do something like:
    //static_assert(std::is_same<>::value, "blah")
    //if you know something about the types.
};

template<typename... Args>
struct convert_in_tuple< variadic_typedef<Args...> > {
    //use Args normally
    typedef std::tuple<Args...> type;
};

typedef variadic_typedef<int, float> myTypes;
typedef convert_in_tuple<myTypes>::type int_float_tuple; //compiles
//typedef convert_in_tuple<int, float>::type int_float_tuple; //doesn't compile

int main() {}