Question

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

-edit-

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; } 

Answer 1

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

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; } }; 

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; };  

and MyClass actually looks like:

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

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; }