問題描述

我想讓這個專門的不改變主.是否可以根據(jù)其基類專門化某些東西?我希望如此.

I want to make this specialized w/o changing main. Is it possible to specialize something based on its base class? I hope so.

-編輯-

我將有幾個繼承自 SomeTag 的類.我不想為他們每個人編寫相同的專業(yè).

I'll have several classes that inherit from SomeTag. I don't want to write the same specialization for each of them.

class SomeTag {};
class InheritSomeTag : public SomeTag {};

template <class T, class Tag=T>
struct MyClass
{
};

template <class T>
struct MyClass<T, SomeTag>
{
    typedef int isSpecialized;
};

int main()
{
    MyClass<SomeTag>::isSpecialized test1; //ok
    MyClass<InheritSomeTag>::isSpecialized test2; //how do i make this specialized w/o changing main()
    return 0;
}

推薦答案

這篇文章描述了一個巧妙的技巧:http://www.gotw.ca/publications/mxc++-item-4.htm

This article describes a neat trick: http://www.gotw.ca/publications/mxc++-item-4.htm

這是基本思想.您首先需要一個 IsDerivedFrom 類(它提供運(yùn)行時和編譯時檢查):

Here's the basic idea. You first need an IsDerivedFrom class (this provides runtime and compile-time checking):

template<typename D, typename B>
class IsDerivedFrom
{
  class No { };
  class Yes { No no[3]; }; 

  static Yes Test( B* ); // not defined
  static No Test( ... ); // not defined 

  static void Constraints(D* p) { B* pb = p; pb = p; } 

public:
  enum { Is = sizeof(Test(static_cast<D*>(0))) == sizeof(Yes) }; 

  IsDerivedFrom() { void(*p)(D*) = Constraints; }
};

那么你的 MyClass 需要一個潛在的特殊實(shí)現(xiàn):

Then your MyClass needs an implementation that's potentially specialized:

template<typename T, int>
class MyClassImpl
{
  // general case: T is not derived from SomeTag
}; 

template<typename T>
class MyClassImpl<T, 1>
{
  // T is derived from SomeTag
  public:
     typedef int isSpecialized;
}; 

和 MyClass 實(shí)際上看起來像:

and MyClass actually looks like:

template<typename T>
class MyClass: public MyClassImpl<T, IsDerivedFrom<T, SomeTag>::Is>
{
};

那么你的主菜就可以了:

Then your main will be fine the way it is:

int main()
{
    MyClass<SomeTag>::isSpecialized test1; //ok
    MyClass<InheritSomeTag>::isSpecialized test2; //ok also
    return 0;
}

這篇關(guān)于基于繼承類的模板特化的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！