通过枚举模板参数编译时间模板选择

Question

我想根据给定的枚举模板参数（store_type）选择一个类模板。现在我实例化一个使用这个类的类，但它似乎总是尝试实例化这个类的basic_store。

enum store_type 
{ 
    none, 
    basic, 
    lockless, 
}; 

template<class T, store_type S = none, typename = void> 
struct get_store_type 
{ 
}; 

template<class T> 
struct get_store_type<T, basic, 
typename std::enable_if<!std::is_abstract<T>::value>::type> 
{ 
    using store_type = typename basic_store<T>; 
}; 

template<class T> 
struct get_store_type<T, lockless> 
{ 
    using store_type = typename lockless_store<T>; 
}; 

template<typename T, store_type S> 
class client 
{ 
public: 
    using my_store_type = typename get_store_type<T, S>::store_type; 
} 

//Tries to instantiate a basic store... which is not allowed. 
client<SomeAbstractType, lockless> something; 

Answer 1

你忘了在专业化模板参数。

template<class T> struct get_store_type<T, lockless, void > 
                ^^^^ 

以下代码的输出是，和：

#include <iostream> 

enum store_type { none, basic, lockless }; 

template<class T, store_type S = none, typename = void> 
struct get_store_type 
{ int a = 1; }; 

template<class T> 
struct get_store_type<T, basic, typename std::enable_if<!std::is_abstract<T>::value>::type> 
{ int b = 2; }; 

template<class T> 
struct get_store_type<T, lockless, void > 
{ int c = 3; }; 

struct Any{}; 

int main(void) 
{ 
    get_store_type<int> storeA; 
    get_store_type<Any, basic> storeB; 
    get_store_type<int, lockless> storeC; 

    std::cout << storeA.a << std::endl; 
    std::cout << storeB.b << std::endl; 
    std::cout << storeC.c << std::endl;  

    return 0; 
}