Virtual destructors

[Dave]

class base
{
public:
virtual ~base();
virtual void foo();
};

class derived: public base
{
public:
~derived();
void foo();
};

derived::foo() does not need to be explicitly designated virtual (though it
is not an error to do so) - it is virtual by virtue of the fact that there
is a member function of the same signature in class base that is virtual.
Along the same lines, is derived::~derived() virtual because base::~base()
is virtual? Or must we explicitly designate derived::~derived() virtual if
we wish it to be so (say, if because we wish to derive from class derived)?

 

[Gianni Mariani]

class base
{
public:
virtual ~base();
virtual void foo();
};

class derived: public base
{
public:
~derived();
void foo();
};

derived::foo() does not need to be explicitly designated virtual (though it
is not an error to do so) - it is virtual by virtue of the fact that there
is a member function of the same signature in class base that is virtual.
Along the same lines, is derived::~derived() virtual because base::~base()
is virtual? Or must we explicitly designate derived::~derived() virtual if
we wish it to be so (say, if because we wish to derive from class derived)?

in the case above, both derived member functions are virtual.

hence :

base * b = new derived();

delete b; // this calls ~derived() ....

 

[Tim Threlfall]

class base
{
public:
virtual ~base();
virtual void foo();
};

class derived: public base
{
public:
~derived();
void foo();
};

derived::foo() does not need to be explicitly designated virtual (though it
is not an error to do so) - it is virtual by virtue of the fact that there
is a member function of the same signature in class base that is virtual.
Along the same lines, is derived::~derived() virtual because base::~base()
is virtual? Or must we explicitly designate derived::~derived() virtual if
we wish it to be so (say, if because we wish to derive from class derived)?

12.4.7 of the standard states:

A destructor can be declared virtual (10.3) or pure virtual (10.4); if
any objects of that class or any derived class are created in the
program, the destructor shall be defined. If a class has a base class
with a virtual destructor, its destructor (whether user- or implicitly-
declared) is virtual.

 

[lilburne]

derived::foo() does not need to be explicitly designated virtual (though it
is not an error to do so) - it is virtual by virtue of the fact that there
is a member function of the same signature in class base that is virtual.
Along the same lines, is derived::~derived() virtual because base::~base()
is virtual? Or must we explicitly designate derived::~derived() virtual if
we wish it to be so (say, if because we wish to derive from class derived)?

You can't revoke virtual, once a function is declared
virtual all subsequent functions with the same signature are
virtual too.

Whilst you don't have to explicitly declare either
derived::foo() or derived::~derived() virtual, it is polite
to do so. That way readers of derived class don't have to
read back through the class heirarchy to discover which
methods are or are not virtual.