Get base class for a type in class hierarchy

Question

Is is possible to get base class type in a class hierarchy?

For example:

struct A{};
struct B{} : public A;
struct C{} : public B;

I want some template that will have typedef Base<T>::Type inside like this:

Base<A>::Type == A
Base<B>::Type == A
Base<C>::Type == A

Is this possible? What about the case when I have multiple inheritance?

Answer 1

Classes in C++ can have more than one base class, so there's no sense in having a "get me the base" trait.

However, the TR2 additions include new compiler-supported traits std::tr2::bases and std::tr2::direct_bases, which returns an opaque type list of base classes.

I'm not sure whether this will make it into C++14, or whether it'll be released independently, but GCC already seems to support this.

Answer 2

I think std::is_base_of can help you

#include <type_traits>

std::is_base_of<B, D>()

If D is derived from B or if both are the same non-union class, provides the member constant value equal to true. Otherwise value is false.

You can use it to check if a class is base class of another or not :

std::is_base_of<A, A>()   // Base<A>::Type == A

std::is_base_of<A, B>()   // Base<B>::Type == A

std::is_base_of<A, C>()   // Base<C>::Type == A