我试图让用户生成他们自己的存储类型，但我迷失了最后一个模板元函数。
基本存储类:
template < typename Data,
typename P1,
typename P2,
typename P3 >
struct Storage : P1, P2, P3 {};
元函数:
template < template <typename,typename> typename Container,
template <typename...> typename WrapperType,
template <typename> typename Allocator = std::allocator >
struct MetaStorage {
template < template <typename> typename P1,
template <typename> typename P2,
template <typename> typename P3 >
struct With_Policies {
template < typename ... Ts >
struct With_Types {
template < typename T = WrapperType<Ts...>,
typename Data = Container<T, Allocator<T>> >
using type = Storage<Data,
P1<Data>,
P2<Data>,
P3<Data>>;
};
};
};
用例:
template <typename ... T > struct DefaultWrapper {};
template < typename T > struct Policie1 {};
template < typename T > struct Policie3 {};
template < typename T > struct Policie2 {};
struct C1 {};
struct C2 {};
using Sig = MetaStorage<std::vector, DefaultWrapper>::With_Policies<Policie1, Policie2, Policie3>::With_Types<C1, C2>::type;
错误 g++ 6.3:
test.cpp: In function ‘int main()’: test.cpp:137:15: error: invalid use of template-name ‘MetaStorage<std::vector, DefaultWrapper>::With_Policies<Policie1, Policie2, Policie3>::With_Types<C1, C2>::type’ without an argument list using Sig = MetaStorage<std::vector, DefaultWrapper>::With_Policies<Policie1, Policie2, Policie3>::With_Types<C1, C2>::type;
^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ test.cpp:58:37: note: ‘template<class T, class Data> using type = Storage<Data, Policie1<Data>, Policie2<Data>, Policie3<Data> >’ declared here
P3<Data>>;
^
最佳答案
type被声明为别名模板，那么这里需要模板参数。因为声明了默认参数，所以只需添加 <>为此，例如
using Sig = MetaStorage<std::vector, DefaultWrapper>::With_Policies<Policie1, Policie2, Policie3>::With_Types<C1, C2>::type<>;
// ~~
关于c++ - 未为使用默认模板参数声明的别名模板指定模板参数时出错，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/43304715/