C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  Virtual Destructors?          C++Talk.NET Forum Index -> C++ language (comp.lang.c++) View previous topic :: View next topic   Author Message amit_edge@yahoo.com Guest Posted: Fri Apr 29, 2005 6:48 am    Post subject: Virtual Destructors? Can anyone tell me the reason for this behaviour of virtual destructors? #include<stdio.h> class a { public: int i; a() { i=10; } virtual ~a() { printf("nDestroying A"); } }; class b: public a { public: ~b(){ printf("nDestroying B"); } }; void main() { b *b1=new b; a *a1=b1; a1->~a(); //calls Destructor B and A printf("%dn",a1->i); // prints 10 delete b1; //calls Destructor A only!!!! printf("%dn",a1->i); // prints junk } Back to top roberth+news@ifi.uio.no Guest Posted: Fri Apr 29, 2005 7:09 am    Post subject: Re: Virtual Destructors? [email]amit_edge (AT) yahoo (DOT) com[/email] wrote: Quote: Can anyone tell me the reason for this behaviour of virtual destructors? Your program is ill-formed, and when run, (after changing void main to int main, so that it will compile, of course) anything can happen. -- Robert Bauck Hamar Back to top Sumit Rajan Guest Posted: Fri Apr 29, 2005 7:18 am    Post subject: Re: Virtual Destructors? <amit_edge (AT) yahoo (DOT) com> wrote Quote: Can anyone tell me the reason for this behaviour of virtual destructors? #include<stdio.h class a { public: int i; a() { i=10; } virtual ~a() { printf("nDestroying A"); } }; class b: public a { public: ~b(){ printf("nDestroying B"); } }; You want to change this to "int main()". For more details, see: http://www.parashift.com/c++-faq-lite/newbie.html#faq-29.3 Quote: void main() { b *b1=new b; a *a1=b1; a1 and b1 point to the same object. Quote: a1->~a(); //calls Destructor B and A now you destroy the object. Quote: printf("%dn",a1->i); // prints 10 Here, you invoke undefined behaviour (UB) by trying to play around with an object that does not exist. Quote: delete b1; //calls Destructor A only!!!! You delete an object twice! Further UB, but now it really doesn't matter anymore since we're already in the realm of undefined behaviour. http://www.parashift.com/c++-faq-lite/freestore-mgmt.html#faq-16.2 Quote: printf("%dn",a1->i); // prints junk More of the same. UB+anything = UB Regards, Sumit. -- Sumit Rajan <sumit.rajan (AT) gmail (DOT) com> Back to top Sumit Rajan Guest Posted: Fri Apr 29, 2005 7:29 am    Post subject: Re: Virtual Destructors? "Sumit Rajan" <sumit.rajan (AT) gmail (DOT) com> wrote Quote: [email]amit_edge (AT) yahoo (DOT) com[/email]> wrote in message news:1114757285.175718.35040 (AT) f14g2000cwb (DOT) googlegroups.com... Can anyone tell me the reason for this behaviour of virtual destructors? #include<stdio.h class a { public: int i; a() { i=10; } virtual ~a() { printf("nDestroying A"); } }; class b: public a { public: ~b(){ printf("nDestroying B"); } }; You want to change this to "int main()". For more details, see: http://www.parashift.com/c++-faq-lite/newbie.html#faq-29.3 void main() { b *b1=new b; a *a1=b1; a1 and b1 point to the same object. a1->~a(); //calls Destructor B and A now you destroy the object. Oops! The destructor is called but the memory is not released. Regards, Sumit. -- Sumit Rajan <sumit.rajan (AT) gmail (DOT) com> Back to top amit_edge@yahoo.com Guest Posted: Fri Apr 29, 2005 8:40 am    Post subject: Re: Virtual Destructors? That is what I am asking Sumit. OK we make main() return int, now tell me how is it that the destructor is called but the object is not destroyed. Why does "delete b1" then call the destructor of A only?? Destructor B destroyed successfully in the first attempt?? Back to top Jean-Sebastien Samson Guest Posted: Fri Apr 29, 2005 8:48 am    Post subject: Re: Virtual Destructors? Quote: That is what I am asking Sumit. OK we make main() return int, now tell me how is it that the destructor is called but the object is not destroyed. Why does "delete b1" then call the destructor of A only?? Destructor B destroyed successfully in the first attempt?? What Sumit told you is that you delete an object which no longer exists (even if the memory was not released). This is evil and results in undefined behaviour. Your program might also legitimately erase your hard drive and launch a nuclear missile. The standard authorizes this. Therefore, it is impossible to explain to you why you specifically get this result as any result would qualify as undefined behaviour (even working perfectly during all the development phase and blowing latter in the face of your most important customer). -- JS Back to top Peter Koch Larsen Guest Posted: Fri Apr 29, 2005 9:12 am    Post subject: Re: Virtual Destructors? <amit_edge (AT) yahoo (DOT) com> skrev i en meddelelse news:1114764003.430276.247910 (AT) l41g2000cwc (DOT) googlegroups.com... Quote: That is what I am asking Sumit. OK we make main() return int, now tell me how is it that the destructor is called but the object is not destroyed. Why does "delete b1" then call the destructor of A only?? Destructor B destroyed successfully in the first attempt?? delete has a dual purpose: 1) destroy the object by calling the objects destructor 2) releasing the memory. You are only allowed to call a destructor once - calling it more times than one causes undefined behaviour. Also, there is almost never any reason to call the destructor explicitly as you did. What was the purpose of that? /Peter Back to top Sumit Rajan Guest Posted: Fri Apr 29, 2005 9:24 am    Post subject: Re: Virtual Destructors? <amit_edge (AT) yahoo (DOT) com> wrote Quote: That is what I am asking Sumit. OK we make main() return int, now tell me how is it that the destructor is called but the object is not From the C++ Standard, 12.4/13: "Once a destructor is invoked for an object, the object no longer exists; ....." Maybe this will be helpful, too: http://www.parashift.com/c++-faq-lite/dtors.html#faq-11.9 Quote: destroyed. Why does "delete b1" then call the destructor of A only?? As mentioned in Jean-Sebastien Samson's posting, you are now attempting to destroy an object that does not exist. Regards, Sumit. -- Sumit Rajan <sumit.rajan (AT) gmail (DOT) com> Back to top Rolf Magnus Guest Posted: Fri Apr 29, 2005 10:39 am    Post subject: Re: Virtual Destructors? [email]amit_edge (AT) yahoo (DOT) com[/email] wrote: Quote: That is what I am asking Sumit. OK we make main() return int, now tell me how is it that the destructor is called but the object is not destroyed. The object is destroyed. That's what the destructor does. But the memory isn't released. Quote: Why does "delete b1" then call the destructor of A only?? a1 and b1 are just pointers. By writing "a *a1=b1;", you just made a1 point to the very same object that b1 points to. So there are _not_ two object, but still only one. By "a1->~a()", you destroy that object, so now there is no object left. You still have the memory that the object was in, but C++ doesn't allow you to do much with it. Especially not to try to destroy the (now non-existing) object again. "delete b1;", when used correctly, does that. It first destroys the object, then releases the memory it occupied. Quote: Destructor B destroyed successfully in the first attempt?? 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