C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  Placement new return value Goto page 1, 2  Next          C++Talk.NET Forum Index -> C++ Language (Moderated) View previous topic :: View next topic   Author Message James Hopkin Guest Posted: Tue Nov 09, 2004 9:20 pm    Post subject: Placement new return value Dear all Given any type T and a pointer, p, to suitably aligned memory, if I do #include <new> new (p) T; is static_cast<T*>(p) subsequently guaranteed to be a valid pointer to the newed object? I believe it is, but I'm not sure where to look in the standard for confirmation. Thanks in advance James [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Alberto Barbati Guest Posted: Wed Nov 10, 2004 4:26 pm    Post subject: Re: Placement new return value James Hopkin wrote: Quote: Dear all Given any type T and a pointer, p, to suitably aligned memory, if I do #include <new new (p) T; is static_cast subsequently guaranteed to be a valid pointer to the newed object? I believe it is, but I'm not sure where to look in the standard for confirmation. To be precise, it's static_cast<T*>(static_cast<void*>(p)) that is guaranteed to be a valid pointer. The cast to void* is required in case p is a pointer to a type multiply or virtually derived from T. The guarantee comes from §4.10/2 and the fact that that placement form of operator new returns its argument (§18.4.1.3/2). Alberto [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Victor Bazarov Guest Posted: Wed Nov 10, 2004 4:26 pm    Post subject: Re: Placement new return value James Hopkin wrote: Quote: Given any type T and a pointer, p, to suitably aligned memory, if I do #include <new new (p) T; is static_cast subsequently guaranteed to be a valid pointer to the newed object? I believe it is, but I'm not sure where to look in the standard for confirmation. 18.4.1.3 V [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Ron Natalie Guest Posted: Wed Nov 10, 2004 4:28 pm    Post subject: Re: Placement new return value James Hopkin wrote: Quote: Dear all Given any type T and a pointer, p, to suitably aligned memory, if I do #include <new new (p) T; is static_cast subsequently guaranteed to be a valid pointer to the newed object? I believe it is, but I'm not sure where to look in the standard for confirmation. Yes, but that is not the case for new (p) T[n]; The standard doesn't come right out and say it, but the allocator for a scalar new gets the exact size of the requested type. The array new is the array length times the size plus some overhead. See 5.3.4/10 and 12 [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Maxim Yegorushkin Guest Posted: Thu Nov 11, 2004 11:13 am    Post subject: Re: Placement new return value On 9 Nov 2004 16:20:15 -0500, James Hopkin <jhopkin (AT) reflectionsinteractive (DOT) com> wrote: Quote: Given any type T and a pointer, p, to suitably aligned memory, if I do #include <new new (p) T; is static_cast subsequently guaranteed to be a valid pointer to the newed object? I believe it is, but I'm not sure where to look in the standard for confirmation. Only when p has type of void*. Standard guaranties that pointer of any type can be safely converted only to void* and back. -- Maxim Yegorushkin [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top kanze@gabi-soft.fr Guest Posted: Thu Nov 11, 2004 11:15 am    Post subject: Re: Placement new return value [email]jhopkin (AT) reflectionsinteractive (DOT) com[/email] (James Hopkin) wrote in message news:<a0368c9c.0411090317.2517b5f7 (AT) posting (DOT) google.com>... Quote: Given any type T and a pointer, p, to suitably aligned memory, if I do #include <new new (p) T; is static_cast subsequently guaranteed to be a valid pointer to the newed object? Not if T is an array type. Otherwise, I think so, but I'm not too sure where to look either. -- James Kanze GABI Software http://www.gabi-soft.fr Conseils en informatique orientée objet/ Beratung in objektorientierter Datenverarbeitung 9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34 [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Maxim Yegorushkin Guest Posted: Fri Nov 12, 2004 4:05 am    Post subject: Re: Placement new return value On 10 Nov 2004 11:28:48 -0500, Ron Natalie <ron (AT) sensor (DOT) com> wrote: Quote: James Hopkin wrote: Dear all Given any type T and a pointer, p, to suitably aligned memory, if I do #include <new new (p) T; is static_cast subsequently guaranteed to be a valid pointer to the newed object? I believe it is, but I'm not sure where to look in the standard for confirmation. Yes, but that is not the case for new (p) T[n]; How about this: struct T { int n; }; struct U { int m; }; struct V : U, T {}; V v, *p = &v; new (p) T; static_cast<T*>(p); // oops, points to (char*)p + sizeof(int) -- to (uninitialized) T subobject of nonexistent V -- Maxim Yegorushkin [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Thomas Mang Guest Posted: Fri Nov 12, 2004 12:10 pm    Post subject: Re: Placement new return value "Maxim Yegorushkin" <e-maxim (AT) yandex (DOT) ru> schrieb im Newsbeitrag news:opsg8rv5rrti5cme (AT) devlx007 (DOT) ipcb.net... Quote: On 9 Nov 2004 16:20:15 -0500, James Hopkin [email]jhopkin (AT) reflectionsinteractive (DOT) com[/email]> wrote: Given any type T and a pointer, p, to suitably aligned memory, if I do #include <new new (p) T; is static_cast subsequently guaranteed to be a valid pointer to the newed object? I believe it is, but I'm not sure where to look in the standard for confirmation. Only when p has type of void*. Standard guaranties that pointer of any type can be safely converted only to void* and back. If I read the Standard correctly, the conversions from base <-> derived are also guaranteed to work. Now take this example: class some_class{}; int main() { void* Memory = new char[10 * sizeof(some_class)]; some_class * Buffer = static_cast<some_class*>(Memory); } The array-new expression yields a void * (a), that is then converted to a char* (b) and then back to void* (c). Now this void* is converted into some_class* (d). a -> c should be fine, what about d? From one point of view, we are only converting a void * to a some_class* (c -> d). However, this is a subsequence of the conversion from a char* to a some_class* (b -> d). Is this little programm now defined or not? Another question regarding the same idiom of allocating a buffer in a char[]: Is this line: some_class * Buffer = (some_class*) new char[10 * sizeof(some_class)]; equivalent to 1) some_class * Buffer = static_cast<some_class*>(static_cast<void*>(new char[10 * sizeof(some_class)])); or equivalent to 2) some_class * Buffer = reinterpret_cast<some_class*>(new char[10 * sizeof(some_class)]); ???? In case 1), the same issue as raised above would apply. In case 2), the result is unspecified anyways. Thomas [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Branimir Maksimovic Guest Posted: Fri Nov 12, 2004 3:08 pm    Post subject: Re: Placement new return value Ron Natalie <ron (AT) sensor (DOT) com> wrote Quote: James Hopkin wrote: Dear all Given any type T and a pointer, p, to suitably aligned memory, if I do #include <new new (p) T; is static_cast subsequently guaranteed to be a valid pointer to the newed object? I believe it is, but I'm not sure where to look in the standard for confirmation. Yes, but that is not the case for new (p) T[n]; The standard doesn't come right out and say it, but the allocator for a scalar new gets the exact size of the requested type. The array new is the array length times the size plus some overhead. See 5.3.4/10 and 12 This rises another question. When using void* operator new[](size_t s)throw(std::bad_alloc); compiler does the trick of calculating *that* overhead; but, in case of placement version there is no portable way of knowing how much bytes to reserve for it. So we must supply another version of placement new in order to avoid buffer overuns: void* operator new[](size_t required_size,size_t real_size,void* p) throw(std::bad_alloc) { if(real_size return p; } This is the question: Is placement version of array new completely useless? :) Greetings, Bane. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Maxim Yegorushkin Guest Posted: Sun Nov 14, 2004 8:40 pm    Post subject: Re: Placement new return value Thomas Mang wrote: Quote: Only when p has type of void*. Standard guaranties that pointer of any type can be safely converted only to void* and back. If I read the Standard correctly, the conversions from base <-> derived are also guaranteed to work. Yes, but in the case with multiple base classes a compiler may adjust a pointer, so that the cast result is not equal to the casted pointer. Please, see my previous posting for a sample. -- Maxim Yegorushkin [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Thomas Richter Guest Posted: Wed Nov 17, 2004 2:13 am    Post subject: Re: Placement new return value Hi, Quote: When using void* operator new[](size_t s)throw(std::bad_alloc); compiler does the trick of calculating *that* overhead; but, in case of placement version there is no portable way of knowing how much bytes to reserve for it. Right, but you don't need to. The compiler computes that, and includes it in the size argument. Which is in bytes. Quote: So we must supply another version of placement new in order to avoid buffer overuns: void* operator new[](size_t required_size,size_t real_size,void* p) throw(std::bad_alloc) { if(real_size return p; } No. What for? The operator new[] only provides raw memory. You ain't supposed to know how many elements you need to reserve memory for. Only the size of the memory is required to get it. Quote: This is the question: Is placement version of array new completely useless? No. So long, Thomas [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Branimir Maksimovic Guest Posted: Thu Nov 18, 2004 12:42 am    Post subject: Re: Placement new return value Thomas Richter <thor (AT) cleopatra (DOT) math.tu-berlin.de> wrote Quote: Hi, When using void* operator new[](size_t s)throw(std::bad_alloc); compiler does the trick of calculating *that* overhead; but, in case of placement version there is no portable way of knowing how much bytes to reserve for it. Right, but you don't need to. The compiler computes that, and includes it in the size argument. Which is in bytes. Of course, but I'm simply asking this: class A{ public: A(): puff_(0xdeadbeef) { } int puff()const{ return puff_; } ~A(){std::cout< private: int puff_; }; int main() { // Question: will this segfault on some systems? char buf[10*sizeof(A)]; A* abuf = new ((void*)buf) A[10]; for(size_t i = 0;i<10;++i) { std::cout< abuf[i].~A(); } } Quote: So we must supply another version of placement new in order to avoid buffer overuns: void* operator new[](size_t required_size,size_t real_size,void* p) throw(std::bad_alloc) { if(real_size return p; } No. What for? The operator new[] only provides raw memory. You ain't supposed to know how many elements you need to reserve memory for. Only the size of the memory is required to get it. Number of elements is not a question. I mean this: char buf[num_of_el*sizeof(T)]; new (sizeof(buf), (void*)buf) T[num_of_el]; // supplied version can throw std::bad_alloc Greetings, Bane. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Thomas Richter Guest Posted: Fri Nov 19, 2004 12:57 pm    Post subject: Re: Placement new return value Hi, Quote: Right, but you don't need to. The compiler computes that, and includes it in the size argument. Which is in bytes. Of course, but I'm simply asking this: /* snip */ Quote: // Question: will this segfault on some systems? char buf[10*sizeof(A)]; A* abuf = new ((void*)buf) A[10]; for(size_t i = 0;i<10;++i) { std::cout< abuf[i].~A(); } Ok, I see what you're aiming at. The point is that you cannot know the byte size of the required memory in advance, thus you cannot use placement array new for a placement allocation. (-; Thus, the above could really segfault. But that doesn't render placement array new useless. It just has other uses. I guess the main problem is its name. Placement new can do a lot of wonderful things besides "placing" objects into memory. It should rather be called "new with parameters". A typical use for the placement new argument would be that of a memory pool handle from which the memory has to be taken for the allocation. In the above example, you wouldn't need to use placement new in first place. Just allocating the "A" class objects one after another at (buf + i * sizeof(A)) would also do that without the problem at hand (keeping care of proper exception handling as well). So long, Thomas [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Thomas Mang Guest Posted: Fri Nov 19, 2004 4:06 pm    Post subject: Re: Placement new return value "Branimir Maksimovic" <bmaxa (AT) volomp (DOT) com> schrieb im Newsbeitrag news:88cdcbb3.0411171007.543e21e2 (AT) posting (DOT) google.com... Quote: Thomas Richter <thor (AT) cleopatra (DOT) math.tu-berlin.de> wrote Hi, When using void* operator new[](size_t s)throw(std::bad_alloc); compiler does the trick of calculating *that* overhead; but, in case of placement version there is no portable way of knowing how much bytes to reserve for it. Right, but you don't need to. The compiler computes that, and includes it in the size argument. Which is in bytes. Of course, but I'm simply asking this: class A{ public: A(): puff_(0xdeadbeef) { } int puff()const{ return puff_; } ~A(){std::cout< private: int puff_; }; int main() { // Question: will this segfault on some systems? char buf[10*sizeof(A)]; A* abuf = new ((void*)buf) A[10]; for(size_t i = 0;i<10;++i) { std::cout< abuf[i].~A(); } Of course it can because it is undefined behaviour. First, you assign a value to an int the Standard does not guarantee it can represent. Second, you ignore alignment. Thomas [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Branimir Maksimovic Guest Posted: Sat Nov 20, 2004 6:14 pm    Post subject: Re: Placement new return value "Thomas Mang" <nospam (AT) nospam (DOT) invalid> wrote Quote: "Branimir Maksimovic" <bmaxa (AT) volomp (DOT) com> schrieb im Newsbeitrag news:88cdcbb3.0411171007.543e21e2 (AT) posting (DOT) google.com... Thomas Richter <thor (AT) cleopatra (DOT) math.tu-berlin.de> wrote in message news:<cnct35$ne1$1 (AT) mamenchi (DOT) zrz.TU-Berlin.DE>... Hi, When using void* operator new[](size_t s)throw(std::bad_alloc); compiler does the trick of calculating *that* overhead; but, in case of placement version there is no portable way of knowing how much bytes to reserve for it. Right, but you don't need to. The compiler computes that, and includes it in the size argument. Which is in bytes. Of course, but I'm simply asking this: class A{ public: A(): puff_(0xdeadbeef) { } int puff()const{ return puff_; } ~A(){std::cout< private: int puff_; }; int main() { // Question: will this segfault on some systems? char buf[10*sizeof(A)]; A* abuf = new ((void*)buf) A[10]; for(size_t i = 0;i<10;++i) { std::cout< abuf[i].~A(); } Of course it can because it is undefined behaviour. First, you assign a value to an int the Standard does not guarantee it can represent. Second, you ignore alignment. You got it wrong. It is not undefined behavior because of that. First check out 4.7, it says value is implementation defined if it cannot be represented in destination signed type. Alse see 5.3.4/10 for explanation why alignment is not an issue here. The other Thomas got it right. Problem is that you cannot portably know *right* size without digging out parameter from operator new[]. Greetings, Bane. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. 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