C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  How does the destructor get called?          C++Talk.NET Forum Index -> C++ Language (Moderated) View previous topic :: View next topic   Author Message Dave Guest Posted: Thu Dec 25, 2003 7:10 am    Post subject: How does the destructor get called? Hello all, When I run the program below, I see this output: base::base() 42 derived::derived() 42 derived::~derived() 42 base::~base() 42 This is, of course, correct. However, I don't understand how the derived class destructor is getting called. I have a const reference to *base* that is referring to a derived object. Is this not akin to having a pointer to base pointing to a derived object, where you would expect the derived class destructor to get called only if the base class destructor is virtual? Thanks, Dave #include <iostream> class base { public: base(int d): base_data(d) { std::cout << "base::base() " << base_data << std::endl; } base(const base &other): base_data(other.base_data) { std::cout << "base::base(const base &) " << base_data << std::endl; } ~base() { std::cout << "base::~base() " << base_data << std::endl; } private: int base_data; }; class derived: public base { public: derived(int d): base(d), derived_data(d) { std::cout << "derived::derived() " << derived_data << std::endl; } derived(const derived &other): base(other), derived_data(other.derived_data) { std::cout << "derived::derived(const derived &) " << derived_data << std::endl; } ~derived() { std::cout << "derived::~derived() " << derived_data << std::endl; } private: int derived_data; }; int main() { const base &r = derived(42); (void) r; // Suppress compiler warning } [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Jonathan Turkanis Guest Posted: Thu Dec 25, 2003 2:39 pm    Post subject: Re: How does the destructor get called? "Dave" <better_cs_now (AT) yahoo (DOT) com> wrote: Quote: When I run the program below, I see this output: base::base() 42 derived::derived() 42 derived::~derived() 42 base::~base() 42 This is, of course, correct. However, I don't understand how the derived class destructor is getting called. I have a const reference to *base* that is referring to a derived object. int main() { const base &r = derived(42); (void) r; // Suppress compiler warning } The expression 'derived(42)' creates a temporary object of type derived which is bound to the reference r. This temporary object persists for the lifetime of r (12.2 paragraph 5). When the reference goes out of scope, the temporary's destructor is called (12.2 paragraph 3). Jonathan [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Francis Glassborow Guest Posted: Thu Dec 25, 2003 2:40 pm    Post subject: Re: How does the destructor get called? In message <vukl5ff4htil16 (AT) news (DOT) supernews.com>, Dave <better_cs_now (AT) yahoo (DOT) com> writes Quote: When I run the program below, I see this output: base::base() 42 derived::derived() 42 derived::~derived() 42 base::~base() 42 This is, of course, correct. However, I don't understand how the derived class destructor is getting called. I have a const reference to *base* that is referring to a derived object. Is this not akin to having a pointer to base pointing to a derived object, where you would expect the derived class destructor to get called only if the base class destructor is virtual? Thanks, Dave #include instrumented base and derived classes snipped int main() { const base &r = derived(42); At this point you have a temporary derived whose life has been extended by binding it to a const reference. However the compiler is responsible for the eventual destruction of that temporary when r goes out of scope; it fulfils its responsibility. Quote: (void) r; // Suppress compiler warning } Note that this program exposes the lack of virtual dtors: int main(){ base * b_ptr = new derived; // the use of 'new' makes the programmer responsible for the // correct destruction of the object delete b_ptr; // and the compiler will use the static (compile time) type of b_ptr // to select the base dtor because of the lack of virtual } -- Francis Glassborow ACCU Author of 'You Can Do It!' see http://www.spellen.org/youcandoit or http://www.robinton.demon.co.uk Happy Xmas, Hanukkah, Yuletide, Winter/Summer Solstice to all. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Siemel Naran Guest Posted: Thu Dec 25, 2003 2:43 pm    Post subject: Re: How does the destructor get called? "Dave" <better_cs_now (AT) yahoo (DOT) com> wrote in message Quote: base::base() 42 derived::derived() 42 derived::~derived() 42 base::~base() 42 This is, of course, correct. However, I don't understand how the derived class destructor is getting called. I have a const reference to *base* that is referring to a derived object. Is this not akin to having a pointer to base pointing to a derived object, where you would expect the derived class destructor to get called only if the base class destructor is virtual? int main() { const base &r = derived(42); (void) r; // Suppress compiler warning } The expression "derived(42)" creates a temporary derived object which must eventually be destroyed. The lhs "const base& r =" binds the temporary to a reference. The next line (void)r has no effect. At the end of the function block, variable r is not destroyed because it is a reference. However, the temporary derived(42) is destroyed, and that explains the output you see. Had variable been an object, things would be different. int main() { const base r = derived(42); (void) r; // Suppress compiler warning } The output of this program should I think be base::base() 42 derived::derived() 42 base::base(const base&) 42 // from copying 'r' derived::~derived() 42 // from deleting temporary base::~base() 42 // finishing deleting temporary base::~base() 42 // from deleting 'r' -- +++++++++++ Siemel Naran [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Conrad Weyns Guest Posted: Thu Dec 25, 2003 5:38 pm    Post subject: Re: How does the destructor get called? "Dave" <better_cs_now (AT) yahoo (DOT) com> wrote Quote: Hello all, When I run the program below, I see this output: base::base() 42 derived::derived() 42 derived::~derived() 42 base::~base() 42 This is, of course, correct. However, I don't understand how the derived class destructor is getting called. I have a const reference to *base* that is referring to a derived object. Is this not akin to having a pointer to base pointing to a derived object, where you would expect the derived class destructor to get called only if the base class destructor is virtual? The object is not getting destroyed through the reference - as would be to case when deleting a derived through a pointer to base. In you example, it is a temporary object (a complete derived) that is getting auto-detroyed. Having "assigned" it to a const base& only forces its life time to be extended to the lifetime of the reference. When a reference goes out of scope, there can be no implicit destruction of the object it is referencing. Conrad Weyns. Quote: Thanks, Dave [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Ulrich Eckhardt Guest Posted: Thu Dec 25, 2003 5:38 pm    Post subject: Re: How does the destructor get called? Dave wrote: Quote: When I run the program below, I see this output: base::base() 42 derived::derived() 42 derived::~derived() 42 base::~base() 42 int main() { const base &r = derived(42); (void) r; // Suppress compiler warning } You should insert another few debug-staements into main to also see when exactly the thing is destroyed. Anyhow, the rule is that if a temporary is bound to a const reference, its lifetime is extended to the scope of the reference. And that is all you see. If you hadn't bound it to a const reference, it would have been just the same, only the 'when' would have changed. For some clever uses of this, see e.g. http://www.cuj.com/documents/s=8250/cujcexp2106alexandr/ Uli -- Questions ? see C++-FAQ Lite: http://parashift.com/c++-faq-lite/ first ! [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Peter Koch Larsen Guest Posted: Thu Dec 25, 2003 5:39 pm    Post subject: Re: How does the destructor get called? "Dave" <better_cs_now (AT) yahoo (DOT) com> skrev i en meddelelse news:vukl5ff4htil16 (AT) news (DOT) supernews.com... Quote: Hello all, When I run the program below, I see this output: base::base() 42 derived::derived() 42 derived::~derived() 42 base::~base() 42 This is, of course, correct. However, I don't understand how the derived class destructor is getting called. I have a const reference to *base* that is referring to a derived object. Is this not akin to having a pointer to base pointing to a derived object, where you would expect the derived class destructor to get called only if the base class destructor is virtual? Thanks, Dave [snip] Quote: int main() { const base &r = derived(42); (void) r; // Suppress compiler warning } Hi Dave Nothing weird is going on. When r (a REFERENCE to a class) is bound to the temporary derived object, the rule is that the lifetime of that object should be as long as the lifetime of r. Please note that there is no base object around, except for the one that is part of the derived object. Kind regards Peter [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. 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