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;
}
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;
}