Why doesn't compile? I would have thought the dtor of Base would call
Derived::f()
struct Base
{
virtual void f() = 0;
virtual ~Base(){f();}
};
struct Derived : public Base
{
virtual void f(){cout << "\nIn Derived f";}
};
int main()
{
Derived d;
return 0;
} 6 2242
Uzytkownik "TheFerryman" <fe***@onthenet.com> napisal w wiadomosci
news:25********************************@4ax.com... Why doesn't compile? I would have thought the dtor of Base would call Derived::f()
struct Base { virtual void f() = 0;
virtual ~Base(){f();} };
struct Derived : public Base { virtual void f(){cout << "\nIn Derived f";} };
int main() { Derived d;
return 0; }
Destructor of Base cannot call Derived::f(), because Derived does not exist
at this point. At Base::~Base, d is of class Base, because Derived
destructor was already executed. In general, virtual function call mechanism
does not work during construction/destruction phase. Imagine what would
happen if Base::~Base actually called Derived::f() - a method of object
would be called after its destructor has already executed! It would break
important C++ 'contract', which states that no operation can be performed on
object before constructor or after destructor.
Best regards,
Marcin
Uzytkownik "TheFerryman" <fe***@onthenet.com> napisal w wiadomosci
news:25********************************@4ax.com... Why doesn't compile? I would have thought the dtor of Base would call Derived::f()
struct Base { virtual void f() = 0;
virtual ~Base(){f();} };
struct Derived : public Base { virtual void f(){cout << "\nIn Derived f";} };
int main() { Derived d;
return 0; }
Destructor of Base cannot call Derived::f(), because Derived does not exist
at this point. At Base::~Base, d is of class Base, because Derived
destructor was already executed. In general, virtual function call mechanism
does not work during construction/destruction phase. Imagine what would
happen if Base::~Base actually called Derived::f() - a method of object
would be called after its destructor has already executed! It would break
important C++ 'contract', which states that no operation can be performed on
object before constructor or after destructor.
Best regards,
Marcin
On Mon, 5 Apr 2004 13:58:24 +0200, "Marcin Kalicinski"
<ka****@poczta.onet.pl> wrote: Uzytkownik "TheFerryman" <fe***@onthenet.com> napisal w wiadomosci news:25********************************@4ax.com.. . Why doesn't compile? I would have thought the dtor of Base would call Derived::f()
struct Base { virtual void f() = 0;
virtual ~Base(){f();} };
struct Derived : public Base { virtual void f(){cout << "\nIn Derived f";} };
int main() { Derived d;
return 0; }
Destructor of Base cannot call Derived::f(), because Derived does not exist at this point. At Base::~Base, d is of class Base, because Derived destructor was already executed. In general, virtual function call mechanism does not work during construction/destruction phase. Imagine what would happen if Base::~Base actually called Derived::f() - a method of object would be called after its destructor has already executed! It would break important C++ 'contract', which states that no operation can be performed on object before constructor or after destructor.
Best regards, Marcin
Ah yes! I see that now. Thanks for your explanation.
On Mon, 5 Apr 2004 13:58:24 +0200, "Marcin Kalicinski"
<ka****@poczta.onet.pl> wrote: Uzytkownik "TheFerryman" <fe***@onthenet.com> napisal w wiadomosci news:25********************************@4ax.com.. . Why doesn't compile? I would have thought the dtor of Base would call Derived::f()
struct Base { virtual void f() = 0;
virtual ~Base(){f();} };
struct Derived : public Base { virtual void f(){cout << "\nIn Derived f";} };
int main() { Derived d;
return 0; }
Destructor of Base cannot call Derived::f(), because Derived does not exist at this point. At Base::~Base, d is of class Base, because Derived destructor was already executed. In general, virtual function call mechanism does not work during construction/destruction phase. Imagine what would happen if Base::~Base actually called Derived::f() - a method of object would be called after its destructor has already executed! It would break important C++ 'contract', which states that no operation can be performed on object before constructor or after destructor.
Best regards, Marcin
Ah yes! I see that now. Thanks for your explanation.
Marcin Kalicinski wrote: Destructor of Base cannot call Derived::f(), because Derived does not exist at this point. At Base::~Base, d is of class Base, because Derived destructor was already executed. In general, virtual function call mechanism does not work during construction/destruction phase.
Actually, it does work. You just have to remember that the object isn't
of the Derived class anymore in the Base destructor. There is a
difference. Think of the following example:
#include <iostream>
class Base
{
public:
virtual void foo() { std::cout << "Base\n"; }
void x () { foo(); }
};
class Derived : public Base
{
public:
void foo() { std::cout << "Derived\n"; }
~Derived() { x(); }
};
class Another : public Derived
{
public:
void foo() { std::cout << "Another\n"; }
};
int main()
{
Another a;
}
The output of this program would be "Base" if polymorphism wouldn't work
in the destructor, but it does print "Derived".
Marcin Kalicinski wrote: Destructor of Base cannot call Derived::f(), because Derived does not exist at this point. At Base::~Base, d is of class Base, because Derived destructor was already executed. In general, virtual function call mechanism does not work during construction/destruction phase.
Actually, it does work. You just have to remember that the object isn't
of the Derived class anymore in the Base destructor. There is a
difference. Think of the following example:
#include <iostream>
class Base
{
public:
virtual void foo() { std::cout << "Base\n"; }
void x () { foo(); }
};
class Derived : public Base
{
public:
void foo() { std::cout << "Derived\n"; }
~Derived() { x(); }
};
class Another : public Derived
{
public:
void foo() { std::cout << "Another\n"; }
};
int main()
{
Another a;
}
The output of this program would be "Base" if polymorphism wouldn't work
in the destructor, but it does print "Derived". This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics
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