基于适用于所有派生策略的基本策略获取模板类专业化

我有一些源自基本策略的策略。某些类专门用于派生策略，而其他类仅专门用于基本策略并且可以与所有派生策略一起使用。

我遇到的问题是代码重复太多（主要是类本身的构造函数和一些样板代码）。下面的代码可以更好地解释我的意思：

struct BasePolicy {};
struct DerivedPolicy1 : public BasePolicy {};
struct DerivedPolicy2 : public BasePolicy {};
//... more policies deriving from BasePolicy (or other BasePolicies)
struct AnotherPolicy {};

template <typename T>
struct Foo;

// This struct can work for all BasePolicy types which includes all derivations
// of it (but doesn't because it is specialized for BasePolicy only)
template<>
struct Foo<BasePolicy>
{
  //... many constructors along with code
};

template<>
struct Foo<AnotherPolicy>
{
  //... more code
};

/* Would like to avoid the following code as it duplicates the above when it 
   comes to constructors and other things such as typedefs */
//template<>
//struct Foo<DerivedPolicy1> : Foo<BasePolicy>
//{
//  //... same constructors as Foo<BasePolicy>
//};
//
//template<>
//struct Foo<DerivedPolicy2> : Foo<BasePolicy>
//{
//  //... same constructors as Foo<BasePolicy>
//};

int main()
{
  // would like this to compile without uncommenting the commented out code and
  // without having the client (i.e. the line below) to somehow get the base
  // type of the policy (although if it can be done transparently, that will
  // work)
  Foo<DerivedPolicy1> a; 
};

有没有办法让派生策略被专门用于基本策略的类接受？我希望客户不要做任何额外的事情。

以下不是有效的 C++ 代码，但我希望发生这样的事情（如果您记住上面的代码）：

template<>
struct Foo<BasePolicy | DerivedPolicy1 | DerivedPolicy2>
{
  //... many constructors along with code
};

这是 SFINAE 的一个案例。

template< typename Policy, ...some_condition... >
struct Foo<Policy>
{
 ...
};

您应该准确地确定 some_condition 是什么。您可以指定 Policy 派生自 BasePolicy：

template< typename Policy, enable_if< is_base<BasePolicy, Policy> > >

或者您可以明确列出允许的策略：

template< typename Policy,
           enable_if_c <is_same<Policy, BasePolicy>::value || 
                         is_same<Policy, DerivedPolicy1>::value> || 
                         ...whatever...
                        >
         >