default virtual d'tor - c++

Question

Let us assume I have two classes:
class Base{};
class Derived: public Base{};
none has d'tor, in this case if I declare about variables:
Base b;
Derived d;
my compiler will produce for me d'tors, my question is, the default d'tors of the b and d will be virtual or not?

Answer 1

my question is, the d'tors of the b and d will be virtual or not
No, they won't. If you want a virtual destructor, you will have to define your own, even if its implementation is exactly the same as that which would be supplied by the compiler:
class Base {
public:
virtual ~Base() {}
};

Answer 2

The destructors of Base and Derived will not be virtual. To make a virtual destructor you need to mark it up explicitly:
struct Base
{
virtual ~Base() {}
};
Actually there's now only one reason to use virtual destructors. That is to shut up the gcc warning: "class 'Base' has virtual functions but non-virtual destructor". As long as you always store your allocated objects in a shared_ptr, then you really don't need a virtual destructor. Here's how:
#include <iostream> // cout, endl
#include <memory> // shared_ptr
#include <string> // string
struct Base
{
virtual std::string GetName() const = 0;
};
class Concrete : public Base
{
std::string GetName() const
{
return "Concrete";
}
};
int main()
{
std::shared_ptr<Base> b(new Concrete);
std::cout << b->GetName() << std::endl;
}
The shared_ptr will clean up correctly, without the need for a virtual destructor. Remember, you will need to use the shared_ptr though!
Good luck!

Answer 3

my question is, the d'tors of the b and d will be virtual or not
Short answer : Nopes!

Answer 4

They will NOT be virtual.However, if you declared(and defined) a virtual dtor in Base, then the derived's dtor would be automatically virtual. HTH.

Answer 5

How can they be virtual unless you explicitly make them as virtual

Answer 6

Just to add one more example to Daniel Lidström's answer
As long as you always store your allocated objects in a shared_ptr, then you really don't need a virtual destructor.
If one uses a shared_ptr like this:
std::shared_ptr<Base> b(new Concrete);
Then the Concrete destructor and the Base destructor are called on destruction of the object.
If one uses a shared_ptr like this:
Base* pBase = new Concrete;
std::shared_ptr<Base> b(pBase);
Then only the Base destructor is called on destruction of the object.
This is an example
#include <iostream> // cout, endl
#include <memory> // shared_ptr
#include <string> // string
struct Base
{
virtual std::string GetName() const = 0;
~Base() { std::cout << "~Base\n"; }
};
struct Concrete : public Base
{
std::string GetName() const
{
return "Concrete";
}
~Concrete() { std::cout << "~Concrete\n"; }
};
int main()
{
{
std::cout << "test 1\n";
std::shared_ptr<Base> b(new Concrete);
std::cout << b->GetName() << std::endl;
}
{
std::cout << "test 2\n";
Base* pBase = new Concrete;
std::shared_ptr<Base> b(pBase);
std::cout << b->GetName() << std::endl;
}
}