Question

当我偶然发现这个问题时，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;

当我尝试输入模板参数包时，GCC 4.5.0给出了一个错误。

基本上，我想将参数包“存储”在typedef中，而不需要解压缩。可能吗？如果没有，是否有一些理由不允许这样做？

Answer 1

另一种比Ben更通用的方法如下：

#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()
{}

Answer 2

我认为它不被允许的原因是它会很混乱，你可以解决它。您需要使用依赖项反转并将存储参数包的结构化为工厂模板，以便将该参数包应用于另一个模板。

有些事情：

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;

Answer 3

我发现Ben Voigt的想法在我自己的努力中非常有用。我稍微对它进行了修改，使其不仅仅是元组。对于读者来说，这可能是一个明显的修改，但值得展示：

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源于这样一个事实，即该类型现在使用前一个列表进行实例化。

Answer 4

这是GManNickG整洁的部分专精技巧的变体。没有委托，您可以通过要求使用variadic_typedef结构来获得更多的类型安全性。

#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() {}

是否可以“存储”模板参数包而不扩展它？

4 个答案: