问题描述
当模板参数派生自特定基类时,我有一个只有静态成员 list
的类,如果该参数不是从基类派生的,则程序不应该 编译,因为静态成员不存在。
我现在遇到的问题是我不知道如何初始化 list
。
这是代码:
#include <type_traits>
template<typename T,size_t MaxSubscribers_>
struct Event
{
using Args_T = T;
static constexpr size_t MaxSubscribers = MaxSubscribers_;
};
class EventManager
{
public:
template<typename T,size_t MaxSubscribers>
struct subscriberMap
{
template<typename E,typename Valid = void>
struct events {};
template<typename E>
struct events<E,typename std::enable_if<std::is_base_of<Event<T,MaxSubscribers>,E>::value>::type>
{
static typename E::Args_T list[E::MaxSubscribers];
};
};
};
// my attempt,does not compile because list doesn't exsist
template<typename T,size_t MaxSubscribers>
template<typename E,typename Valid>
typename E::Args_T EventManager::subscriberMap<T,MaxSubscribers>::events<E,Valid>::list[E::MaxSubscribers];
解决方法
解决方案是您必须重复定义中的确切模板参数列表。
template<typename T,size_t MaxSubscribers>
template<typename E>
typename E::Args_T EventManager::subscriberMap<T,MaxSubscribers>::events<E,typename std::enable_if<std::is_base_of<Event<T,MaxSubscribers>,E>::value>::type>::list[E::MaxSubscribers];
,
从 c++17 开始,您可以内联声明静态成员:
static inline typename E::Args_T list[E::MaxSubscribers]{};
// ^^^^^^ ^^
// maybe a meaningful default value