Question

I am trying to write a template that will extract the base type of a boost::shared_ptr.

I wrote this template:

template<typename T>
struct ExtractBaseType;

template<typename T>
struct ExtractBaseType<boost::shared_ptr<T> > 
{
    typedef T type;
};

it works fine for a plain shared_ptr. This:

struct A
{
};

ExtractBaseType<boost::shared_ptr<A> >::type  a_thing;
std::cout << typeid(a_thing).name() << std::endl;

prints “1A”.

However, this doesn’t compile:

struct B : boost::shared_ptr<A>
{
};

ExtractBaseType<B>::type  b_thing;

The compiler complains that ExtractBaseType is undefined.

Why so? And how would this be done?

Answer 1

it doesn’t work because you are matching shared_ptr not B. you need to match derived of shared_ptr.

template<typename T, class = void>
struct ExtractBaseType;

template<class C>
struct ExtractBaseType<
    C, typename enable_if<
           boost::is_base_of<shared_ptr<typename T::element_type>, T>::value
       >::type
    > 
{
    typedef typename T::element_type type;
};

^ didn’t test, but the main idea is there

Good question. That said, inheriting from shared_ptr seems ugly.