C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  Is sizeof() evauated at compile time ? Goto page 1, 2, 3, 4, 5  Next          C++Talk.NET Forum Index -> C++ Language (Moderated) View previous topic :: View next topic   Author Message dragoncoder Guest Posted: Thu Dec 01, 2005 4:35 pm    Post subject: Is sizeof() evauated at compile time ? Okay, here is the code. #include <cstdlib> #include <iostream> int main() { int* ptr = NULL; // Do I have to put std::NULL ? std::cout << sizeof (*ptr) << std::endl; return 0; } Questions: 1. Do I have to write std::NULL ? As I know all standard library names are defined in std namespace. 2. Is sizeof () evaluated at compile time ? I want to know if the code produces undefined behaviour. Thanks [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Chris Uzdavinis Guest Posted: Fri Dec 02, 2005 12:33 am    Post subject: Re: Is sizeof() evauated at compile time ? Quote: Questions: 1. Do I have to write std::NULL ? As I know all standard library names are defined in std namespace. No. Macros are not in namespaces. Quote: 2. Is sizeof () evaluated at compile time ? I want to know if the code produces undefined behaviour. sizeof() is evaluated at compile time, and its evaluation does not consider values, only types. Your code is fine. One common "trick" people do is take the sizeof() a function invocation expression, which gives the size of the return type of the function without actually calling it. -- Chris [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top peter steiner Guest Posted: Fri Dec 02, 2005 12:33 am    Post subject: Re: Is sizeof() evauated at compile time ? dragoncoder wrote: Quote: Okay, here is the code. #include #include int main() { int* ptr = NULL; // Do I have to put std::NULL ? std::cout << sizeof (*ptr) << std::endl; return 0; } Questions: 1. Do I have to write std::NULL ? As I know all standard library names are defined in std namespace. NULL is defined to be a macro, thus namespaces do not apply. (C.2.2.3) Quote: 2. Is sizeof () evaluated at compile time ? I want to know if the code produces undefined behaviour. yes, sizeof is (kind of) evaluated at compile time. if sizeof is given an expression it just computes the resulting object size. the expression is not evaluated in the sense of pointer dereference that you are worrying about. your sizeof operator call is legal and directly translates to sizeof(int). (5.3.3) -- peter [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Razzer Guest Posted: Fri Dec 02, 2005 12:37 am    Post subject: Re: Is sizeof() evauated at compile time ? dragoncoder wrote: Quote: Okay, here is the code. #include #include int main() { int* ptr = NULL; // Do I have to put std::NULL ? std::cout << sizeof (*ptr) << std::endl; return 0; } Questions: 1. Do I have to write std::NULL ? As I know all standard library names are defined in std namespace. No. NULL is a marco. It *cannot* have a std:: prefix. Quote: 2. Is sizeof () evaluated at compile time ? I want to know if the code produces undefined behaviour. Yes. Your code is well-formed. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top berad Guest Posted: Fri Dec 02, 2005 12:38 am    Post subject: Re: Is sizeof() evauated at compile time ? no you don't have to put std:null, and no its evaluated at runtime, at least that would make sense to me [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Bob Hairgrove Guest Posted: Fri Dec 02, 2005 12:52 am    Post subject: Re: Is sizeof() evauated at compile time ? On 1 Dec 2005 11:35:55 -0500, "dragoncoder" <pktiwary (AT) gmail (DOT) com> wrote: Quote: Okay, here is the code. #include #include int main() { int* ptr = NULL; // Do I have to put std::NULL ? std::cout << sizeof (*ptr) << std::endl; return 0; } Questions: 1. Do I have to write std::NULL ? As I know all standard library names are defined in std namespace. NULL is a macro defined by some compilers. It is not part of the ANSI standard C++ library, and therefore it is not in namespace std. You should actually write: int *ptr = 0; instead of NULL. Bjarne Stroustrup, among others, recommends this (see: "The C++ Programming Language", chapter 5, section 5.1.1). Quote: 2. Is sizeof () evaluated at compile time ? I want to know if the code produces undefined behaviour. sizeof() is indeed evaluated at compile time. Because the compiler knows that ptr points to int, it knows the size of *ptr even if it hasn't yet been set to point to a valid object. Somewhere in the standard it says that no expressions in the sizeof() statement are actually evaluated, hence there are no side effects by writing *ptr here. I might be wrong, but I believe the behavior of your code is well-defined. -- Bob Hairgrove [email]NoSpamPlease (AT) Home (DOT) com[/email] [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Clark S. Cox III Guest Posted: Fri Dec 02, 2005 12:53 am    Post subject: Re: Is sizeof() evauated at compile time ? On 2005-12-01 11:35:55 -0500, "dragoncoder" <pktiwary (AT) gmail (DOT) com> said: Quote: Okay, here is the code. #include <cstdlib #include int main() { int* ptr = NULL; // Do I have to put std::NULL ? std::cout << sizeof (*ptr) << std::endl; return 0; } Questions: 1. Do I have to write std::NULL ? As I know all standard library names are defined in std namespace. No. NULL is a macro. Macros are simple text substitution, and don't obey scoping rules. Quote: 2. Is sizeof () evaluated at compile time ? Yes it is. Quote: I want to know if the code produces undefined behaviour. sizeof only relies on the *type* of its operand, never its *value*. sizeof doesn't evaluate its operand. your program will produce the same output as: #include int main() { std::cout << sizeof (int) << std::endl; return 0; } -- Clark S. Cox, III [email]clarkcox3 (AT) gmail (DOT) com[/email] [ 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: Fri Dec 02, 2005 12:54 am    Post subject: Re: Is sizeof() evauated at compile time ? "dragoncoder" <pktiwary (AT) gmail (DOT) com> writes: Quote: #include <cstdlib #include std::endl is declared in as well. Quote: int main() { int* ptr = NULL; // Do I have to put std::NULL ? std::cout << sizeof (*ptr) << std::endl; return 0; } Questions: 1. Do I have to write std::NULL ? As I know all standard library names are defined in std namespace. No. NULL is a macro and therefore doesn't belong to a namespace. Quote: 2. Is sizeof () evaluated at compile time? Yes. Quote: I want to know if the code produces undefined behaviour. I don't think it does. The only problem is the lacking #inclusion of to work fine (if <iostream> happens to #include <ostream>). [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top gottlobfrege@gmail.com Guest Posted: Fri Dec 02, 2005 12:54 am    Post subject: Re: Is sizeof() evauated at compile time ? Quote: Is sizeof() evauated at compile time ? Yes. dragoncoder wrote: Quote: Okay, here is the code. #include #include int main() { int* ptr = NULL; // Do I have to put std::NULL ? std::cout << sizeof (*ptr) << std::endl; return 0; } Questions: 1. Do I have to write std::NULL ? As I know all standard library names are defined in std namespace. IMO, just use 0 instead of NULL. Quote: 2. Is sizeof () evaluated at compile time ? I want to know if the code produces undefined behaviour. Interesting question. Technically, by the wording of the standard, it might be arguable, but I will bet that you are fine for all implementations. Quote: Thanks [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Thomas Tutone Guest Posted: Fri Dec 02, 2005 12:55 am    Post subject: Re: Is sizeof() evauated at compile time ? dragoncoder wrote: Quote: #include #include int main() { int* ptr = NULL; // Do I have to put std::NULL ? std::cout << sizeof (*ptr) << std::endl; return 0; } Questions: 1. Do I have to write std::NULL ? As I know all standard library names are defined in std namespace. No. NULL is a macro. Quote: 2. Is sizeof () evaluated at compile time ? Yes. Quote: I want to know if the code produces undefined behaviour. No, I believe it's well defined - it should output whatever the size of an int is on your system. Best regards, Tom [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Clark S. Cox III Guest Posted: Fri Dec 02, 2005 11:43 am    Post subject: Re: Is sizeof() evauated at compile time ? On 2005-12-01 19:33:52 -0500, "peter steiner" <pnsteiner (AT) gmail (DOT) com> said: Quote: dragoncoder wrote: 2. Is sizeof () evaluated at compile time ? I want to know if the code produces undefined behaviour. yes, sizeof is (kind of) evaluated at compile time. What do you mean by "kind of"? I thought that, in C++ (and in C, if you discount C99's VLA's) sizeof always results in a compile-time constant. -- Clark S. Cox, III [email]clarkcox3 (AT) gmail (DOT) com[/email] [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Clark S. Cox III Guest Posted: Fri Dec 02, 2005 11:44 am    Post subject: Re: Is sizeof() evauated at compile time ? On 2005-12-01 19:54:02 -0500, Thomas Maeder <maeder (AT) glue (DOT) ch> said: Quote: "dragoncoder" <pktiwary (AT) gmail (DOT) com> writes: #include <cstdlib #include std::endl is declared in as well. Every single code example using endl in the standard includes <iostream>, and not <ostream> -- Clark S. Cox, III [email]clarkcox3 (AT) gmail (DOT) com[/email] [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Jack Klein Guest Posted: Fri Dec 02, 2005 11:46 am    Post subject: Re: Is sizeof() evauated at compile time ? On 1 Dec 2005 19:52:40 -0500, Bob Hairgrove <invalid (AT) bigfoot (DOT) com> wrote in comp.lang.c++.moderated: [snip] Quote: Questions: 1. Do I have to write std::NULL ? As I know all standard library names are defined in std namespace. NULL is a macro defined by some compilers. It is not part of the ANSI standard C++ library, and therefore it is not in namespace std. You NULL is a macro defined by specified headers in every single conforming C and C++ compiler. Macros defined in headers are considered part of the library. [snip] -- Jack Klein Home: http://JK-Technology.Com FAQs for comp.lang.c http://www.eskimo.com/~scs/C-faq/top.html comp.lang.c++ http://www.parashift.com/c++-faq-lite/ alt.comp.lang.learn.c-c++ http://www.contrib.andrew.cmu.edu/~ajo/docs/FAQ-acllc.html [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Daniel Krügler Guest Posted: Fri Dec 02, 2005 11:49 am    Post subject: Re: Is sizeof() evauated at compile time ? Razzer wrote: Quote: dragoncoder wrote: Okay, here is the code. #include <cstdlib #include int main() { int* ptr = NULL; // Do I have to put std::NULL ? std::cout << sizeof (*ptr) << std::endl; return 0; } Questions: 1. Do I have to write std::NULL ? As I know all standard library names are defined in std namespace. No. NULL is a marco. It *cannot* have a std:: prefix. That is not true in general. Assert also is a macro, but depending on your compiler compliance and whether you included versus <assert.h> std::assert can be wellformed. The same reasoning applies to std::errno and others. The reason is clear, of course: Whether a namespace-specifier is needed, depends on the way of macrofying.. Greetings from Bremen, Daniel Krügler [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Hrayr BABAJANYAN Guest Posted: Fri Dec 02, 2005 11:50 am    Post subject: Re: Is sizeof() evauated at compile time ? Hi! 1. No! NULL is a macro defined like #define NULL 0 // or #define NULL ((void*)0) // macros are evaluated by preprocessor // and the name resolutions at copile time // so there is no std::NULL (in std namespace) 2. Yes! operator sizeof is being evaluated at compile time, it computes something like 'typeof' for the expression inside ()s (the type of the expression), and then computes the size of that type. well the code listed looks ok, no undefined behaviour, the size of expression *ptr is defined at compile time (depends on the platform you use). Cheers! Hrayr [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. 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