变分变分模板
Variadic variadic template templates
我目前正在与以下代码作斗争,其目的是实现可变可变模板:
template
<
template <typename... HeadArgs> class Head,
template <typename... TailArgs> class...
>
struct join<Head<typename HeadArgs...>, Head<typename TailArgs...>...>
{
typedef Head<typename HeadArgs..., typename TailArgs......> result;
};
理想情况下,我可以使用这个模板元函数来实现以下功能:
template <typename...> struct obj1 {};
template <typename...> struct obj2 {};
typedef join
<
obj1<int, int, double>,
obj1<double, char>,
obj1<char*, int, double, const char*>
>::result new_obj1;
typedef join
<
obj2<int, int, double>,
obj2<double, char>,
obj2<char*, int, double, const char*>
>::result new_obj2;
/* This should result in an error, because there are
different encapsulating objects
typedef join
<
obj1<int, int, double>,
obj1<double, char>,
obj2<char*, int, double, const char*>
>::result new_obj;
*/
以上的输出将有望以template<int, int, double, double, char, char*, int, double, const char*> struct new_obj[1|2] {};
的形式创建new_obj1
和new_obj2
我在Windows上使用的是gcc 4.6.2,它输出一个"‘…’之前的预期参数包",用于扩展"Head<typename TailArgs...>
…"。
此错误可在gcc 4.5.1中重现。
试试这样的东西:
template <typename...> struct join;
template <template <typename...> class Tpl,
typename ...Args1,
typename ...Args2>
struct join<Tpl<Args1...>, Tpl<Args2...>>
{
typedef Tpl<Args1..., Args2...> type;
};
template <template <typename...> class Tpl,
typename ...Args1,
typename ...Args2,
typename ...Tail>
struct join<Tpl<Args1...>, Tpl<Args2...>, Tail...>
{
typedef typename join<Tpl<Args1..., Args2...>, Tail...>::type type;
};
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