C ++ Microsoft：如何将uuid / guid与模板专门化相关联 [英] C++ Microsoft: How to associate uuid/guid with template specialization

问题描述

我想将uuid / guid与模板专用化关联。

以下代码可用于将uuid与非模板界面class，struct）：

  __ interface __declspec（uuid（CECA446F-2BE6-4AAC-A117-E395F27DF1F8））ITest {
 virtual void Test（）= 0; 
}; 
 
 GUID guid = __uuidof（ITest）; // OK 
  

现在我有一个模板界面

  template< class T> __interface ITemplateTest {
 virtual void Test（T t）= 0; 
}; 
  

，我想进行以下工作：

  GUID templateGuid = __uuidof（ITemplateTest< float>）; 
  

不能将__declspec（uuid（...））添加到模板类定义。这是显而易见的，因为接口的不同专业化需要不同的uuids。所以我试图通过以下方式将uuid与模板专业化关联：

 模板< __interface __declspec（uuid（CF4AB938-8CE0-4AB7-A56C-0253B6018C26））ITemplateTest< float> ;;不幸的是，这不工作（__uuidof（。）失败，没有GUID已关联与此对象）。
 
 
 
我的问题有没有解决方案？
 
 
 
 更完整的示例在稍后添加一些答案后添加： 
 
 
 
假设我有一个适用于不同数据类型的复杂算法。为了简单起见，这种复杂算法是对数字进行平方。我想只实现我的算法一次，所以模板是要走的路。让我们进一步假设我想使用接口，因为我的应用程序使用COM。 
 
 
 
这里是一个接口和一个模板化的对象实现这个接口：
  template< class T> __interface ITemplateTest {
 virtual T Square（T t）= 0; 
}; 
 
 template< class T> class CTemplateImplementation：public ITemplateTest< T> {
 public：
 virtual T Square（T t）{return t * t; }; 
}; 
  
这样可以执行
  CTemplateImplementation< double> xDouble; 
 CTemplateImplementation< float> xFloat; 
 CTemplateImplementation< int> xInt; 
 
 std :: cout<< xDouble.Square（5。）<< std :: endl 
<< xFloat.Square（5.0f）<< std :: endl 
<< xInt.Square（5）<< std :: endl; 
  
 
 
现在让我们进一步假设我有另一个模板对象，也实现了一个非常复杂的算法，使用CTemplateImplementation：
  template< class T> class CSquareAndAddOne {
 private：
 CTemplateImplementation< T> m_squarer; 
 public：
 T SquareAndAddOne（T t）{return m_squarer.Square（t）+ T（1）; } 
}; 
  
此对象现在可以以同样的方式使用：
  CSquareAndAddOne< double> yDouble; 
 CSquareAndAddOne< float> yFloat; 
 CSquareAndAddOne< int> yInt; 
 
 std :: cout<< yDouble.SquareAndAddOne（5。）<< std :: endl 
<< yFloat.SquareAndAddOne（5.0f）< std :: endl 
<< yInt.SquareAndAddOne（5）<< std :: endl; 
  
问题出现，当CSquareAndAddOne :: SquandAndAddOne想使用CTemplateImplementation专用化的__uuid。尝试以下操作：
  T SquareAndAddOne（T t）{
 GUID guid = __uuidof（m_multiplier）; 
 return m_squarer.Square（t）+ T（1）; 
} 
  
这不再工作，因为没有GUID与m_multiplier 。那么如何在不重复代码的情况下为CTemplateImplementation的不同实现分配GUID（在这种情况下为三个）？你能提供一个完整的解决方案吗？ 
不可能从不同类型的CTemplateImplementation派生不同的类，因为在CSquareAndAddOne中使用正确类型的专门化不能用模板参数控制。
解决方案
我试过一些明显的实例化魔法，它的工作原理（至少VS2008 SP1）：
  template< class T> __interface ITemplateTest {void Test（T t）; }; //接口声明
 
 template __interface ITemplateTest< float> ;; //显式实例化
 
 template __interface __declspec（uuid（CF4AB938-8CE0-4AB7-A56C-0253B6018C26）ITemplateTest< float> ;; //使用uuid关联重复显式实例化
 
 GUID guid = __uuidof（ITemplateTest< float>）; // Enjoy！:-）
  
最初的问题是 __ declspec 试图在实际实例化之前将guid指定给类特殊化。
 
I'd like to associate a uuid/guid with a template specialization.

The following code can be used to associate a uuid with a non-template interface (class, struct):
__interface __declspec(uuid("CECA446F-2BE6-4AAC-A117-E395F27DF1F8")) ITest {
    virtual void Test() = 0;
};

GUID guid = __uuidof(ITest);   // OK
Now I have a templated interface
template<class T> __interface ITemplateTest {
    virtual void Test(T t) = 0;
    };
and I'd like to make the following work:
GUID templateGuid = __uuidof(ITemplateTest<float>);
It is not possible to add the __declspec(uuid(...)) to the template class definition. This is obvious, since different specializations of the interface need different uuids. So I tried to associate the uuid with a template specialization in the following way:
template<> __interface __declspec(uuid("CF4AB938-8CE0-4AB7-A56C-0253B6018C26")) ITemplateTest<float>;
Unfortunately, this doesn't work either (__uuidof(.) fails with "no GUID has been associated with this object").

Is there any solution to my problem?

Here is a more complete example added later after some answers:

Let's say I have a complex algorithm that works on different data types. This "complex" algorithm is squaring a number for the sake of simplicity. I'd like to implement my algorithm only once, so templates is the way to go. Let's further assume that I want to use interfaces, because my app makes use of COM. 

So here is an interface and a templated object that implements this interface:
template<class T> __interface ITemplateTest {
    virtual T Square(T t) = 0;
    };

template<class T> class CTemplateImplementation : public ITemplateTest<T> {
    public:
        virtual T Square(T t) { return t * t; };
    };
This allows to do something like 
CTemplateImplementation<double> xDouble;
CTemplateImplementation<float>  xFloat;
CTemplateImplementation<int>    xInt;

std::cout << xDouble.Square(5.) << std::endl
          << xFloat.Square(5.0f) << std::endl
          << xInt.Square(5) << std::endl;
Now let's further assume that I have another template object, also implementing a very "complex" algorithm, that makes use of CTemplateImplementation:
template<class T> class CSquareAndAddOne {
    private:
        CTemplateImplementation<T> m_squarer;
    public:
        T SquareAndAddOne(T t) { return m_squarer.Square(t) + T(1); }
    };
This object can now be used in the same way:
CSquareAndAddOne<double> yDouble;
CSquareAndAddOne<float>  yFloat;
CSquareAndAddOne<int>    yInt;

std::cout << yDouble.SquareAndAddOne(5.) << std::endl
          << yFloat.SquareAndAddOne(5.0f) << std::endl
          << yInt.SquareAndAddOne(5) << std::endl;
The problem arises, when CSquareAndAddOne::SquandAndAddOne wants to make use of the __uuid of a CTemplateImplementation specialization. Try the following:
    T SquareAndAddOne(T t) {
        GUID guid = __uuidof(m_multiplier);
        return m_squarer.Square(t) + T(1);
        }
This doesn't work any more, because there is no GUID assiciated with m_multiplier. So how can I assign guids to the (three in this case) different implementations of CTemplateImplementation without duplicating code? Can you provide a complete solution?
Deriving different classes from CTemplateImplementation for different types is not possible, since usage of the correctly typed specialization in CSquareAndAddOne can not be controlled with a template argument any more.
解决方案
I tried some explicit instantiation magic and it works OK (at least with VS2008 SP1):
template<class T> __interface ITemplateTest { void Test(T t); }; // Interface declaration

template __interface ITemplateTest<float>; // Explicit instantiation

template __interface __declspec(uuid("CF4AB938-8CE0-4AB7-A56C-0253B6018C26") ITemplateTest<float>; // Repeat explicit instantiation with uuid association

GUID guid = __uuidof(ITemplateTest<float>); // Enjoy! :-)
Looks like initial problem was that __declspec tried to assign guid to class specialization before it was actually instantiated.

                        
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