是否可以基于模板类型参数的嵌套typedef的存在来专门化模板定义？ [英] Is it possible to specialize a template definition based on the existence of a nested typedef of a template type parameter?

问题描述

我有一个模板， template< typename T>类包装器，我想专门化基于 typename T :: context_type 的存在。如果声明 typename T :: context_type ，则 wrapper 实例化的构造函数和赋值运算符重载应该接受强制的 typename T :: context_type 参数。另外， wrapper< T> 对象将在成员数据中存储上下文。如果 typename T :: context_type 不存在，则 wrapper< T> 的构造函数和赋值运算符重载将采用少一个参数，并且不会有其他数据成员。

这是否可能？我可以得到下面的代码来编译，而不改变 config1 ， config2 和 main（）？

  #include< iostream> 
 
 template< typename T，bool context_type_defined = true> 
 class wrapper 
 {
 public：
 typedef typename T :: context_type context_type; 
 
 private：
 context_type ctx; 
 
 public：
 wrapper（context_type ctx_）
：ctx（ctx_）
 {
 std :: cout< T :: context_type exists。 << std :: endl; 
} 
}; 
 
 template< typename T> 
 class wrapper< T，false> 
 {
 public：
 wrapper（）{
 std :: cout< T :: context_type不存在。 << std :: endl; 
} 
}; 
 
 class config1 {
 public：
 typedef int context_type; 
}; 
 
 class config2 {
 public：
}; 
 
 int main（）
 {
 wrapper< config1> w1（0）; 
 wrapper< config2> w2; 
} 
  

解决方案

我已经通过使用一些元编程技巧在过去实现这样的行为。基本构建块是：

BOOST_MPL_HAS_XXX_TRAIT_DEF ，定义一个元函数谓词，如果参数是类类型，并且具有给定名称的嵌套类型（在您的情况下为context_type）。

http://www.boost.org/doc/libs/1_47_0/libs/mpl /doc/refmanual/has-xxx-trait-def.html

Boost.EnableIf ，基于先前定义的特征定义专业化。

http://www.boost.org/libs/utility/enable_if.html 请参阅3.1启用模板类专业化

---

请注意，您可能能够直接使用SFINAE工作，类似这样的工作：

  template< typename T，typename Context = void> 
 class wrapper {...}; //基本定义
 
 template<类型名T> 
 class wrapper< T，typename voif_mfn < typename T :: context_type> :: type> {...}; // Specialization 
  

然而，我喜欢基于traits的解决方案的表达能力， p>

I have a template, template <typename T> class wrapper, that I would like to specialize based on the existence of typename T::context_type. If typename T::context_type is declared, then the constructors and assignment operator overloads of the wrapper<T> instantiation should accept a mandatory typename T::context_type parameter. Additionally, wrapper<T> objects would store "context" in the member data. If typename T::context_type does not exist, then the constructors and assignment operator overloads of wrapper<T> would take one less parameter and there would be no additional data member.

Is this possible? Can I get the following code to compile without changing the definitions of config1, config2, and main()?

#include <iostream>

template <typename T, bool context_type_defined = true>
class wrapper
{
public:
    typedef typename T::context_type context_type;

private:
    context_type ctx;

public:
    wrapper(context_type ctx_)
        : ctx(ctx_)
    {
        std::cout << "T::context_type exists." << std::endl;
    }
};

template <typename T>
class wrapper<T, false>
{
public:
    wrapper() {
        std::cout << "T::context_type does not exist." << std::endl;
    }
};

class config1 {
public:
    typedef int context_type;
};

class config2 {
public:
};

int main()
{
    wrapper<config1> w1(0);
    wrapper<config2> w2;
}

解决方案

Yes, it is possible. I have implemented such behavior in the past by using some metaprogramming tricks. The basic build blocks are:

BOOST_MPL_HAS_XXX_TRAIT_DEF, to define a metafunction predicate that will evaluate to a true type if the argument is of class type and has a nested type with a given name (context_type in your case).

http://www.boost.org/doc/libs/1_47_0/libs/mpl/doc/refmanual/has-xxx-trait-def.html

Boost.EnableIf, to define the specializations based on the previously defined trait.

http://www.boost.org/libs/utility/enable_if.html # See 3.1 Enabling template class specializations

---

Note that you may be able to get that behavior working directly with SFINAE, something like this may work:

template< typename T, typename Context = void >
class wrapper { ... }; // Base definition

template< typename T >
class wrapper< T, typename voif_mfn< typename T::context_type >::type > { ... }; // Specialization

However, I like the expressiveness of the solution based on traits and enable if.

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