基于所有派生策略的基本策略工作获取模板类专业化

Question

我有从基本策略派生的策略。有些课程专门针对派生策略，而其他课程仅专门针对基本策略，可以与所有衍生策略配合使用。

我遇到的问题是代码重复太多（主要是构造函数和类本身的一些锅炉代码）。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; 
}; 

派生策略是否有任何方法可以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。

下是无效的C++代码，但我想这样的事情发生（如果你把上面的代码记）：

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

Answer 1

这是SFINAE的情况下。

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

您应该完全决定什么是some_condition。您可以指定政策从BasePolicy派生：

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

或者你也可以列出允许的政策明确：

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