Template specialization based on inherit class - c++

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;
}

Answer 2

And the short version now, 2014, using C++-11:
#include <type_traits>
struct SomeTag { };
struct InheritSomeTag : SomeTag { };
template<typename T, bool = std::is_base_of<SomeTag, T>::value>
struct MyClass { };
template<typename T>
struct MyClass<T, true> {
typedef int isSpecialized;
};
int main() {
MyClass<SomeTag>::isSpecialized test1; /* ok */
MyClass<InheritSomeTag>::isSpecialized test2; /* ok */
}

Answer 3

Well, the article in the answer above appeared in February 2002. While it works, today we know there are better ways. Alternatively, you can use enable_if:
template<bool C, typename T = void>
struct enable_if {
typedef T type;
};
template<typename T>
struct enable_if<false, T> { };
template<typename, typename>
struct is_same {
static bool const value = false;
};
template<typename A>
struct is_same<A, A> {
static bool const value = true;
};
template<typename B, typename D>
struct is_base_of {
static D * create_d();
static char (& chk(B *))[1];
static char (& chk(...))[2];
static bool const value = sizeof chk(create_d()) == 1 &&
!is_same<B volatile const,
void volatile const>::value;
};
struct SomeTag { };
struct InheritSomeTag : SomeTag { };
template<typename T, typename = void>
struct MyClass { /* T not derived from SomeTag */ };
template<typename T>
struct MyClass<T, typename enable_if<is_base_of<SomeTag, T>::value>::type> {
typedef int isSpecialized;
};
int main() {
MyClass<SomeTag>::isSpecialized test1; /* ok */
MyClass<InheritSomeTag>::isSpecialized test2; /* ok */
}

Answer 4

In your case, the only way that I see would be to explicitly specialize MyClass for InheritSomeTag. However, the SeqAn paper proposes a mechanism called “template sublassing” that does what you want – albeit with a different inheritance syntax, so the code isn't compatible with your current main function.
// Base class
template <typename TSpec = void>
class SomeTag { };
// Type tag, NOT part of the inheritance chain
template <typename TSpec = void>
struct InheritSomeTag { };
// Derived class, uses type tag
template <typename TSpec>
class SomeTag<InheritSomeTag<TSpec> > : public SomeTag<void> { };
template <class T, class Tag=T>
struct MyClass { };
template <class T, typename TSpec>
struct MyClass<T, SomeTag<TSpec> >
{
typedef int isSpecialized;
};
int main()
{
MyClass<SomeTag<> >::isSpecialized test1; //ok
MyClass<SomeTag<InheritSomeTag<> > >::isSpecialized test2; //ok
}
This certainly looks strange and is very cumbersome but it allows a true inheritance mechanism with polymorphic functions that is executed at compile time. If you want to see this in action, have a look at some SeqAn examples.
That being said, I believe that SeqAn is a special case and not many applications would profit from this extremely difficult syntax (deciphering SeqAn-related compiler errors is a real pain in the *ss!)

Answer 5

Using concepts and the requires keyword from C++20 is an even simpler and more expressive way to do this without having to introduce a redundant boolean non-type template parameter like in C++11:
// C++20:
#include <concepts>
#include <iostream>
struct SomeTag { };
struct InheritSomeTag : SomeTag { };
template<typename T>
struct MyClass
{
void Print()
{
std::cout << "Not derived from someTag\n";
}
};
// std::derived_from is a predefined concept already included in the STL
template<typename T>
requires std::derived_from<T, SomeTag>
struct MyClass<T>
{
void Print()
{
std::cout << "derived from someTag\n";
}
};
int main()
{
MyClass<InheritSomeTag> test1;
test1.Print(); // derived from someTag
MyClass<int> test2;
test2.Print(); // Not derived from someTag
// Note how even the base tag itself returns true from std::derived_from:
MyClass<SomeTag> test3;
test3.Print(); // derived from someTag
}