C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  dereferencing boost::shared_ptr<>          C++Talk.NET Forum Index -> C++ language (comp.lang.c++) View previous topic :: View next topic   Author Message Dennis Jones Guest Posted: Thu Jul 28, 2005 12:08 am    Post subject: dereferencing boost::shared_ptr<> Hello, Given something like: boost::shared_ptr<T> t( new T() ); What is the best (correct?) way to dereference the pointer? The following two methods work. Is there a difference? T &rt1 = *t.get(); T &rt2 = *t; And what about getting at the raw pointers: T *pt1 = t.get(); T *pt2 = &*t; Is there an advantage to one or the other? Thanks, - Dennis Back to top Joe Gottman Guest Posted: Thu Jul 28, 2005 1:05 am    Post subject: Re: dereferencing boost::shared_ptr<> "Dennis Jones" <nospam (AT) nospam (DOT) com> wrote Quote: Hello, Given something like: boost::shared_ptr<T> t( new T() ); What is the best (correct?) way to dereference the pointer? The following two methods work. Is there a difference? T &rt1 = *t.get(); T &rt2 = *t; I prefer the second method, because you get to type four fewer characters Also, suppose you have a typedef somewhere that looks like typedef shared_ptr<T> TPtr; In the second case, you maintain the option of changing this to typedef T *TPtr; without having to change any dereferencing code. Quote: And what about getting at the raw pointers: T *pt1 = t.get(); T *pt2 = &*t; I definitely prefer the first case here. What is t.get() == 0. Then the second case invokes undefined behavior by dereferencing a null pointer. For example, the boost implementation of shared_ptr asserts that get() != 0 inside its implementations of operator*() and operator->(). Joe Gottman Back to top Dennis Jones Guest Posted: Thu Jul 28, 2005 1:20 am    Post subject: Re: dereferencing boost::shared_ptr<> "Joe Gottman" <jgottman (AT) carolina (DOT) rr.com> wrote Quote: What is the best (correct?) way to dereference the pointer? The following two methods work. Is there a difference? T &rt1 = *t.get(); T &rt2 = *t; I prefer the second method, because you get to type four fewer characters Also, suppose you have a typedef somewhere that looks like typedef shared_ptr<T> TPtr; Thank you. I agree with your assessment. I was just a little surprised that the second one compiled and worked. I thought (perhaps naively) it looked like I was dereferencing the shared_ptr instead of the underlying raw pointer. But now that I look at the header file, I see that operator *() is defined to return a reference to the pointee. Quote: And what about getting at the raw pointers: T *pt1 = t.get(); T *pt2 = &*t; I definitely prefer the first case here. What is t.get() == 0. Then the second case invokes undefined behavior by dereferencing a null pointer. For example, the boost implementation of shared_ptr asserts that get() != 0 inside its implementations of operator*() and operator->(). I wouldn't have thought of that -- you're absolutely right. Thanks for your insight, - Dennis Back to top Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year Oldest FirstNewest First         C++Talk.NET Forum Index -> C++ language (comp.lang.c++) All times are GMT Page 1 of 1   Jump to: Select a forum C/C++----------------C LanguageC Language (Moderated)C++ Language (Moderated)C++ language (comp.lang.c++)C++ language, library and standards Non-English C/C++ Newsgroups----------------C++ (German)C++ (French)  You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum Powered by phpBB © 2001, 2006 phpBB Group SEO toolkit © 2004-2006 webmedic.