C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  Why can operator delete be called on a const pointer? Goto page 1, 2  Next          C++Talk.NET Forum Index -> C++ Language (Moderated) View previous topic :: View next topic   Author Message Thomas J Guest Posted: Sun Dec 04, 2005 7:47 pm    Post subject: Why can operator delete be called on a const pointer? Hi! I wondered when i saw a method like this: void MyClass::selfDestruction() const { delete this; } void iAmNotEvil(MyClass const* pMyClass) { p->selfDestruction(); } Why is this possible? When I call a const method i recommend nothing being changend, and in this case the object is being deleted. In my eyes this shouldn't be possible, maybe I overlooked something. Greetings [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Alf P. Steinbach Guest Posted: Mon Dec 05, 2005 12:25 am    Post subject: Re: Why can operator delete be called on a const pointer? * Thomas J: Quote: Hi! I wondered when i saw a method like this: void MyClass::selfDestruction() const { delete this; } void iAmNotEvil(MyClass const* pMyClass) { p->selfDestruction(); } Why is this possible? When I call a const method i recommend nothing being changend, and in this case the object is being deleted. In my eyes this shouldn't be possible, maybe I overlooked something. Destruction is simply construction in reverse. During construction and destruction the object is not const, even if it is const in between these two end-points of the life cycle. Consider struct Ballooba {}; Ballooba const* const p = new Balloba; This is much of the point of constness: a guarantee that at least via this pointer the object won't be modified between initialization and destruction. But if you couldn't call delete with that pointer as argument, you couldn't have a const dynamically allocated object: const would then only be for locals, statics and members -- or for cast-ridden code... -- A: Because it messes up the order in which people normally read text. Q: Why is it such a bad thing? A: Top-posting. Q: What is the most annoying thing on usenet and in e-mail? [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Thomas Maeder Guest Posted: Mon Dec 05, 2005 12:57 am    Post subject: Re: Why can operator delete be called on a const pointer? "Thomas J" <quicix (AT) googlemail (DOT) com> writes: Quote: I wondered when i saw a method like this: void MyClass::selfDestruction() const { delete this; } Why is this possible? The reasoning is that const extends to the lifetime of the object only. The lifetime of an object starts when its constructor has been fully executed and ends when its destruction starts. Quote: When I call a const method i recommend nothing being changend, and in this case the object is being deleted. In my eyes this shouldn't be possible, maybe I overlooked something. You are in good company (not mine, though, but I'm sure you can live with that) AFAICT. What you are asking seems to be a religious question. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top David Abrahams Guest Posted: Mon Dec 05, 2005 10:34 am    Post subject: Re: Why can operator delete be called on a const pointer? "Thomas J" <quicix (AT) googlemail (DOT) com> writes: Quote: Hi! I wondered when i saw a method like this: void MyClass::selfDestruction() const { delete this; } void iAmNotEvil(MyClass const* pMyClass) { p->selfDestruction(); } Why is this possible? For the same reason that a const object can be destroyed when it goes out of scope. Ending the lifetime of an object is not considered a mutating operation. -- Dave Abrahams Boost Consulting www.boost-consulting.com [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Maciej Sobczak Guest Posted: Mon Dec 05, 2005 12:27 pm    Post subject: Re: Why can operator delete be called on a const pointer? Thomas J wrote: Quote: When I call a const method i recommend nothing being changend, and in this case the object is being deleted. This is philosophical, but try to think this way: after deleting, you have absolutely no possibility to check whether anything has changed in the given object (you cannot access it), so you also have no possibility to *prove* that its constness was violated. ;) -- Maciej Sobczak : http://www.msobczak.com/ Programming : http://www.msobczak.com/prog/ [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Dave Harris Guest Posted: Mon Dec 05, 2005 4:52 pm    Post subject: Re: Why can operator delete be called on a const pointer? [email]maeder (AT) glue (DOT) ch[/email] (Thomas Maeder) wrote (abridged): Quote: When I call a const method i recommend nothing being changend, and in this case the object is being deleted. In my eyes this shouldn't be possible, maybe I overlooked something. You are in good company In particular, the C++ Annotated Reference Manual agrees with the original poster. It has some discussion in $5.3.4, which concludes: Without the restriction against deleting pointers to const objects the delete operator would have the effect of implicitly removing the const attribute from a pointer. No operator should have that property. -- Dave Harris, Nottingham, UK. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Sektor van Skijlen Guest Posted: Tue Dec 06, 2005 10:31 am    Post subject: Re: Why can operator delete be called on a const pointer? Dnia 5 Dec 2005 11:52:40 -0500, Dave Harris skrobie: Quote: maeder (AT) glue (DOT) ch (Thomas Maeder) wrote (abridged): When I call a const method i recommend nothing being changend, and in this case the object is being deleted. In my eyes this shouldn't be possible, maybe I overlooked something. You are in good company In particular, the C++ Annotated Reference Manual agrees with the original poster. It has some discussion in $5.3.4, which concludes: Without the restriction against deleting pointers to const objects the delete operator would have the effect of implicitly removing the const attribute from a pointer. No operator should have that property. This seems to be completely wrong for me. The 'delete' operator is not just a normal operator, which receives normal arguments. You could define, for example, operator +, which will take a pointer to appropriate type and its variance must be saved. In case of both 'new' and 'delete' operators, there is no direct passing from the form of its call to its implementation. And it does not change anything that the "operator delete" function receives void* type, which can accept any other data pointer type. Such a conversion does not happen in this case, nor does it happen in case of 'new' operator. Similarly, we could ask, why 'new' operator converts implicitly from void* to pointer to an object. Moreover, why there's no different versions for operator new( void* ) and operator new( const void* ), selected according to if the user requested "new Obj()" or "new const Obj()". Consequences in case of different method of allocation, permissions for these operations etc., I give up to your imaginary. If we want to look deeper into this topic, we could also ask, why you can destroy an object, which's destructor is not const. You can be pushed to an angle seeing that making it so strict would make you also unable to delete the object at all (unless you violate the variance with const_cast). I'd prefer to stop this paranoia. It's best to accept that the 'operator delete' function is called INTERNALLY by the language runtime (from within delete expression) and no pointer conversion rules apply here. This same rule applies to 'new' operator and 'operator new' function as well. Note: you can still use 'operator delete' and 'operator new' functions as well. But in such a case the rules of pointer conversion do apply, as they do in case of any other function called directly (in particular, which's call expression directly refers to definition of its interface). Regards, Sektor -- // _ ___ Michal "Sektor" Malecki <sektor(whirl)kis.p.lodz.pl> \ L_ |/ `| /^ ,() <ethourhs(O)wp.pl> // _ | / _/ / C++ bez cholesterolu: http://www.intercon.pl/~sektor/cbx "I am allergic to Java because programming in Java reminds me casting spells" [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top benben Guest Posted: Tue Dec 06, 2005 10:39 am    Post subject: Re: Why can operator delete be called on a const pointer? Quote: This is philosophical, but try to think this way: after deleting, you have absolutely no possibility to check whether anything has changed in the given object (you cannot access it), so you also have no possibility to *prove* that its constness was violated. ;) That said, it is bad practice to destroy const objects anyway. How do you know the object is heap-allocated. If you can be certain, you should be able to delete the non-const version anyway. Ben [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Sektor van Skijlen Guest Posted: Tue Dec 06, 2005 1:12 pm    Post subject: Re: Why can operator delete be called on a const pointer? Dnia 6 Dec 2005 05:39:58 -0500, benben skrobie: Quote: This is philosophical, but try to think this way: after deleting, you have absolutely no possibility to check whether anything has changed in the given object (you cannot access it), so you also have no possibility to *prove* that its constness was violated. ;) That said, it is bad practice to destroy const objects anyway. How do you know the object is heap-allocated. If you can be certain, you should be able to delete the non-const version anyway. I got a feeling that you are mixing two separate things. If you want to perform 'delete' on a pointer to object, you must ensure that it is appropriate (two things should be taken into consideration: whether the object was allocated with 'new' rather than created by any other way, including automatic on the stack, and whether the information provided for this pointer is enough to destroy it correctly - that is, if it points to object's base class, whether this base class is polymorphic with virtual destructor). The pointer/reference variance (CV-q) does not mean anything in this case. -- // _ ___ Michal "Sektor" Malecki <sektor(whirl)kis.p.lodz.pl> \ L_ |/ `| /^ ,() <ethourhs(O)wp.pl> // _ | / _/ / C++ bez cholesterolu: http://www.intercon.pl/~sektor/cbx "I am allergic to Java because programming in Java reminds me casting spells" [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Martin Bonner Guest Posted: Tue Dec 06, 2005 1:34 pm    Post subject: Re: Why can operator delete be called on a const pointer? benben wrote: Quote: This is philosophical, but try to think this way: after deleting, you have absolutely no possibility to check whether anything has changed in the given object (you cannot access it), so you also have no possibility to *prove* that its constness was violated. ;) That said, it is bad practice to destroy const objects anyway. How do you know the object is heap-allocated. If you can be certain, you should be able to delete the non-const version anyway. Not necessarily. Consider: const int *p = new const int(0); There /is/ no non-const version to delete. (You could use const_cast to get an int* to delete, and I think that is even required by the standard to work. However I would be happy to be proved wrong, and even it is legal it is certainly not moral.) [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Dave Harris Guest Posted: Wed Dec 07, 2005 2:34 pm    Post subject: Re: Why can operator delete be called on a const pointer? [email]martinfrompi (AT) yahoo (DOT) co.uk[/email] (Martin Bonner) wrote (abridged): Quote: Not necessarily. Consider: const int *p = new const int(0); There /is/ no non-const version to delete. In non-template situations you can add one, perhaps wrapped in a smart pointer, like: template <typename T> class Const { T *p; public: Const( T *p = new T() ) : p(p) {} const T *operator->() const { return p; } const T &operator*() const { return *p; } void destroy() { delete p; } }; void test1() { Const<int> p = new int(); p.destroy(); // Replaces delete p. } The wrapper makes sure the non-const version does not escape - the only (purported) non-const operation you can do to Const::p is delete. I suspect the persuasive examples involve template code: template <typename T> void test2( T *p1 ) { T *p2 = new T(); delete p2; } would be an error if instantiated with (const int). Despite that, I personally agree with the ARM: the delete operator conceptually modifies its argument. I'd gladly sacrifice test2() to get const-correctness back. There are other ways to write it. -- Dave Harris, Nottingham, UK. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Dave Harris Guest Posted: Wed Dec 07, 2005 2:36 pm    Post subject: Re: Why can operator delete be called on a const pointer? [email]ethouris (AT) pl (DOT) wp.spamu.lubie.nie.inva[/email]lid (Sektor van Skijlen) wrote (abridged): Quote: In particular, the C++ Annotated Reference Manual agrees with the original poster. It has some discussion in $5.3.4, which concludes: Without the restriction against deleting pointers to const objects the delete operator would have the effect of implicitly removing the const attribute from a pointer. No operator should have that property. This seems to be completely wrong for me. The premise here is that the delete operator effectively modifies the object. Given that, I think the ARM reasoning is sound. (I only quoted a small snippet of it.) Quote: Similarly, we could ask, why 'new' operator converts implicitly from void* to pointer to an object. It can safely do that because it has constructed the object. Indeed, that's its job. Quote: Moreover, why there's no different versions for operator new( void* ) and operator new( const void* ), selected according to if the user requested "new Obj()" or "new const Obj()". Placement new conceptually modifies the memory it's given. So does operator delete(). Even if the object has no user-defined state, the compiler itself may need to modify it, eg to install a vtable pointer or other book-keeping/debugging info. I don't think anyone is asking for const overloads of these in this thread. There's clearly a difference between the delete operator and operator delete. Quote: If we want to look deeper into this topic, we could also ask, why you can destroy an object, which's destructor is not const. Indeed, in a case like: static const Object obj; the const must be removed before the object's destructor is run. However, it doesn't follow that the delete operator is what should do this job. -- Dave Harris, Nottingham, UK. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Seungbeom Kim Guest Posted: Fri Dec 09, 2005 3:13 pm    Post subject: Re: Why can operator delete be called on a const pointer? Dave Harris wrote: Quote: ethouris (AT) pl (DOT) wp.spamu.lubie.nie.invalid (Sektor van Skijlen) wrote (abridged): In particular, the C++ Annotated Reference Manual agrees with the original poster. It has some discussion in $5.3.4, which concludes: Without the restriction against deleting pointers to const objects the delete operator would have the effect of implicitly removing the const attribute from a pointer. No operator should have that property. This seems to be completely wrong for me. The premise here is that the delete operator effectively modifies the object. Given that, I think the ARM reasoning is sound. (I only quoted a small snippet of it.) This is a matter of whether to consider deleting as a kind of modifying. I could argue against it because modifying x means (old x) != (new x) and in the case of delete (new x) doesn't even exist and the inequality doesn't hold. :) (I say "I 'could'" because I mean that it is reasonable to argue so, though I admit that other interpretations are also possible.) Quote: If we want to look deeper into this topic, we could also ask, why you can destroy an object, which's destructor is not const. Indeed, in a case like: static const Object obj; the const must be removed before the object's destructor is run. However, it doesn't follow that the delete operator is what should do this job. Why not? const Object* p = new const Object; // what 'new' does: // (a) creates an Object // (b) labels it const // (c) returns the const object delete p; // what 'delete' does: // (c) receives the const object // (b) removes its const label // (a) destroys the Object This symmetry seems perfectly reasonable to me. -- Seungbeom Kim [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Niklas Matthies Guest Posted: Fri Dec 09, 2005 6:35 pm    Post subject: Re: Why can operator delete be called on a const pointer? On 2005-12-09 15:13, Seungbeom Kim wrote: : Quote: const Object* p = new const Object; // what 'new' does: // (a) creates an Object // (b) labels it const // (c) returns the const object delete p; // what 'delete' does: // (c) receives the const object // (b) removes its const label // (a) destroys the Object This symmetry seems perfectly reasonable to me. One could argue that it is not reasonable for the same reason that normally non-const implicitly converts to const but not vice versa. -- Niklas Matthies [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Paavo Helde Guest Posted: Sat Dec 10, 2005 2:16 am    Post subject: Re: Why can operator delete be called on a const pointer? "Thomas J" <quicix (AT) googlemail (DOT) com> wrote in news:1133719758.592458.17710 @g44g2000cwa.googlegroups.com: Quote: Why is this possible? When I call a const method i recommend nothing being changend, and in this case the object is being deleted. In my eyes this shouldn't be possible, maybe I overlooked something. Yes, you are mixing up the goal and the means to achieve the goal. The const modifier lets you know that you don't have to worry about the object state in the *subsequent* operations on that object - the object content will (or at least should appear to) remain the same. In case of deletion however there is no subsequent operation, thus no worry! hth Paavo [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. 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