编译时间 class 通过枚举模板参数选择模板

Question

我正在尝试 select 基于给定枚举模板参数 (store_type) 的 class 模板。现在我实例化一个使用它的 class，但它似乎总是尝试为此 class.

实例化 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

您忘记了专业化中的 3rd 模板参数。

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

下面代码的输出是1,2和3:

#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;
}

编译时间 class 通过枚举模板参数选择模板

Compile time class template selection by enum template parameter

c++

sfinae

template-meta-programming