Will the template argument's destructor to a templated class be called on deletion?

Question

If you have a templated base class as in the following example:

class A{
   public:
    A();
    virtual ~A();
};

template <class T>
class B : public T
{
  public:
    B();
    virtual ~B();
};

typedef B<A> C;

class D : public C
{
  public:
    D();
    virtual ~D();
};

When you delete an instance of D, will the destructor of A be called?

I'll probably create a test program to find out what happens, but just thinking about it, I wasn't sure what should happen.

Answer 1

When you delete an instance of D, will the destructor of A be called?

Yes. Nothing special here (except you have private access on everything, which means it probably won't compile).

Answer 2

In a bid to get rep from two contradictory answers, given:

#include <iostream>
using namespace std;

class A{
    public:
    A() {}
    virtual ~A() { cout << "~A()\n"; }
};

template <class T>
class B : public T
{
    public:
    B() {}
    virtual ~B() {}
};

typedef B<A> C;

class D : public C
{
    public:
    D() {}
    virtual ~D() {}
};

int main() {
    D d;
}

Then yes, of course A's destructor will be called.