C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  polymorphism - virtual function          C++Talk.NET Forum Index -> C++ language (comp.lang.c++) View previous topic :: View next topic   Author Message al Guest Posted: Wed Dec 31, 2003 3:15 am    Post subject: polymorphism - virtual function class Base { public: virtual void method(); }; class Derive : public Base { public: void method(); }; Base *b = new Base; b->method();//Base::method() called Base *d = new Derive; d->method();//Derive::method() called Why b->method() trigger Base::method() whereas d->method() Derive::method()? Thanks! Back to top Victor Bazarov Guest Posted: Wed Dec 31, 2003 3:20 am    Post subject: Re: polymorphism - virtual function "al" <allin (AT) 168 (DOT) net> wrote... Quote: class Base { public: virtual void method(); }; class Derive : public Base { public: void method(); }; Base *b = new Base; b->method();//Base::method() called Base *d = new Derive; d->method();//Derive::method() called Why b->method() trigger Base::method() whereas d->method() Derive::method()? Because 'b' is a pointer to a complete object of class Base, and 'd' is a pointer to a subobject of a complete object of class Derive. What matters is the type of the object at its creation (the actual, the complete, object). V Back to top Jeff Schwab Guest Posted: Wed Dec 31, 2003 3:26 am    Post subject: Re: polymorphism - virtual function Victor Bazarov wrote: Quote: "al" <allin (AT) 168 (DOT) net> wrote... class Base { public: virtual void method(); }; class Derive : public Base { public: void method(); }; Base *b = new Base; b->method();//Base::method() called Base *d = new Derive; d->method();//Derive::method() called Why b->method() trigger Base::method() whereas d->method() Derive::method()? Because 'b' is a pointer to a complete object of class Base, and 'd' is a pointer to a subobject of a complete object of class Derive. What matters is the type of the object at its creation (the actual, the complete, object). V What Victor said is true for any "virtual" method, including the one you've called here. If you want to change the behavior, just don't include the word "virtual" in the method declaration. Back to top Billy O'Connor Guest Posted: Wed Dec 31, 2003 3:31 am    Post subject: Re: polymorphism - virtual function "al" <allin (AT) 168 (DOT) net> writes: Quote: class Base { public: virtual void method(); }; class Derive : public Base { public: void method(); }; Base *b = new Base; b->method();//Base::method() called Base *d = new Derive; d->method();//Derive::method() called Why b->method() trigger Base::method() whereas d->method() Derive::method()? Well, b is a Base, and d is a Derived. What behavior were you expecting? Back to top al Guest Posted: Wed Dec 31, 2003 4:35 am    Post subject: Re: polymorphism - virtual function Victor Bazarov <v.Abazarov (AT) comAcast (DOT) net> wrote Quote: "al" <allin (AT) 168 (DOT) net> wrote... class Base { public: virtual void method(); }; class Derive : public Base { public: void method(); }; Base *b = new Base; b->method();//Base::method() called Base *d = new Derive; d->method();//Derive::method() called Why b->method() trigger Base::method() whereas d->method() Derive::method()? Because 'b' is a pointer to a complete object of class Base, and 'd' is a pointer to a subobject of a complete object of class Derive. What matters is the type of the object at its creation (the actual, the complete, object). V Thanks! This definitely helps me understand the topic. If removing "virtual" from method() declaraton of class Base, then why d->method() triggers Base::method() since 'd' is a pointer to the actual object of class Derive? Back to top Victor Bazarov Guest Posted: Wed Dec 31, 2003 4:17 pm    Post subject: Re: polymorphism - virtual function "al" <allin (AT) 168 (DOT) net> wrote... Quote: Victor Bazarov <v.Abazarov (AT) comAcast (DOT) net> wrote in message news:ugrIb.706451$Fm2.609354 (AT) attbi_s04 (DOT) .. "al" <allin (AT) 168 (DOT) net> wrote... class Base { public: virtual void method(); }; class Derive : public Base { public: void method(); }; Base *b = new Base; b->method();//Base::method() called Base *d = new Derive; d->method();//Derive::method() called Why b->method() trigger Base::method() whereas d->method() Derive::method()? Because 'b' is a pointer to a complete object of class Base, and 'd' is a pointer to a subobject of a complete object of class Derive. What matters is the type of the object at its creation (the actual, the complete, object). V Thanks! This definitely helps me understand the topic. If removing "virtual" from method() declaraton of class Base, then why d->method() triggers Base::method() since 'd' is a pointer to the actual object of class Derive? No, 'd' is not a pointer to the actual object of class Derive. It is, as I already said, a pointer to a subobject. If you don't use 'virtual' in declaring the member function[s], the binding is static and does not respect the fact that the subobject of type Base is really part of the object created as 'Derive'. The difference, hence, is static binding versus dynamic binding. Victor Back to top Cy Edmunds Guest Posted: Wed Dec 31, 2003 4:31 pm    Post subject: Re: polymorphism - virtual function "al" <allin (AT) 168 (DOT) net> wrote Quote: Victor Bazarov <v.Abazarov (AT) comAcast (DOT) net> wrote in message news:ugrIb.706451$Fm2.609354 (AT) attbi_s04 (DOT) .. "al" <allin (AT) 168 (DOT) net> wrote... class Base { public: virtual void method(); }; class Derive : public Base { public: void method(); }; Base *b = new Base; b->method();//Base::method() called Base *d = new Derive; d->method();//Derive::method() called Why b->method() trigger Base::method() whereas d->method() Derive::method()? Because 'b' is a pointer to a complete object of class Base, and 'd' is a pointer to a subobject of a complete object of class Derive. What matters is the type of the object at its creation (the actual, the complete, object). V Thanks! This definitely helps me understand the topic. If removing "virtual" from method() declaraton of class Base, then why d->method() triggers Base::method() since 'd' is a pointer to the actual object of class Derive? Well, that's the distinction between virtual and not virtual! If not virtual the compiler only takes note of the declared type of the pointer and effectively bit slices down to a Base type (which is OK because a Derive IS a Base). If virtual it looks up the actual type and proceeds accordingly. By the way, Derive::method is virtual also. Once a method is declared virtual you can't make it non-virtual later. -- Cy http://home.rochester.rr.com/cyhome/ Back to top P.V.S.S.VIKAS Guest Posted: Thu Jan 01, 2004 10:29 am    Post subject: Re: polymorphism - virtual function Billy O'Connor <billyoc (AT) gnuyork (DOT) org> wrote Quote: "al" <allin (AT) 168 (DOT) net> writes: class Base { public: virtual void method(); }; class Derive : public Base { public: void method(); }; Base *b = new Base; b->method();//Base::method() called Base *d = new Derive; d->method();//Derive::method() called Why b->method() trigger Base::method() whereas d->method() Derive::method()? Well, b is a Base, and d is a Derived. What behavior were you expecting? these lines have no sense in i'am new to this world.learning to mve around kindly coperate. Back to top Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First         C++Talk.NET Forum Index -> C++ language (comp.lang.c++) All times are GMT Page 1 of 1   Jump to: Select a forum C/C++----------------C LanguageC Language (Moderated)C++ Language (Moderated)C++ language (comp.lang.c++)C++ language, library and standards Non-English C/C++ Newsgroups----------------C++ (German)C++ (French)  You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum Powered by phpBB © 2001, 2006 phpBB Group SEO toolkit © 2004-2006 webmedic.