C++Talk.NET C++ language newsgroups   Archives   FAQ   Search   Memberlist   Usergroups   Register   Profile   Log in to check your private messages   Log in  How to overload operator->*          C++Talk.NET Forum Index -> C++ Language (Moderated) View previous topic :: View next topic   Author Message Tobias Güntner Guest Posted: Tue Jul 20, 2004 7:54 am    Post subject: How to overload operator->* Hi! I guess many people know how to overload operator-> (e.g. in a smart pointer class), but how can one overload operator->* properly? I know it has to return something that can be called like a function and I found a small example at <http://www.briceg.com/ticpp/one/Chapter12.html#Heading362> But how to implement a generic operator->*? At first I tried class Pointer { public: int Work(int i) { return i + 1; } template<typename PMF> class FunctionObject { Pointer* ptr; PMF pmem; public: FunctionObject(Pointer* wp, PMF pmf) : ptr(wp), pmem(pmf) { } template<class T> operator T() const { return ptr->*pmem; } }; template<typename PMF> FunctionObject<PMF> operator->*(PMF pmf) { return FunctionObject<PMF>(this, pmf); } }; int main() { Pointer w; int (Pointer::*pmf)(int) = &Pointer::Work; // error: no match for call to // `(Pointer::FunctionObject<int(Pointer::*)(int)>) (int)' cout << (w->*pmf)(1) << endl; return 0; } but it didn't work (I guess it would have worked if I provided a correct conversion operator, but I don't know which type ptr->*pmem returns. I wonder if it's possible to express this type in C++ at all...) After that I tried the following: class Pointer { public: int Work(int i) { return i + 1; } template<typename R, typename P1> class FunctionObject1 { Pointer* ptr; R (Pointer::*pmem)(P1); public: FunctionObject1(Pointer* wp, R (Pointer::*pmf)(P1)) : ptr(wp), pmem(pmf) { } R operator()(P1 p1) { return (ptr->*pmem)(p1); } }; template<typename R, typename P1> FunctionObject1<R,P1> operator->*(R (Pointer::*pmf)(P1)) { return FunctionObject1<R,P1>(this, pmf); } }; int main() { Pointer w; int (Pointer::*pmf)(int) = &Pointer::Work; cout << (w->*pmf)(1) << endl; return 0; } This worked (g++ 3.3.1), but a) I don't really feel like writing dozens of operator()s with 0..n parameters and b) I don't want too many copies of an object if it's passed by value (although it would probably be possible to solve the latter problem and use a 'const T&' at that places using some sort of traits class). Is there a better/more elegant solution to this problem? Regards, Tobias [ 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: Tue Jul 20, 2004 6:27 pm    Post subject: Re: How to overload operator->* Tobias GГјntner <fatbull (AT) web (DOT) de> wrote: Quote: I guess many people know how to overload operator-> (e.g. in a smart pointer class), but how can one overload operator->* properly? I know it has to return something that can be called like a function and I found a small example at <http://www.briceg.com/ticpp/one/Chapter12.html#Heading362 For data members it returns a reference to the member. struct some { int i_; } s, *ps = &s; int some::*pi = &some::i_; int &ri = ps->*pi; Quote: But how to implement a generic operator->*? You can implement it using boost::function<> and boost::bind: template<class T> struct smart_ptr {/*...*/}; // for one arg member functions template<class T, class R, A0> inline function<R(A0)> operator->*(smart_ptr<T> const& p, R(T::*f)(A0)) { struct invoker { static R invoke(smart_ptr<T> const& p, R(T::*f)(A0), A0 const& a0) { return (p.get()->*f)(a0); } }; return function<R()>(bind(&invoker::invoke, p, f, _1)); } // add versions for N args, member functions, handle reference args -- 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 Joe Guest Posted: Tue Jul 20, 2004 9:35 pm    Post subject: Re: How to overload operator->* "> But how to implement a generic operator->*? Quote: : Do this! ] Several years ago a read an article on the Dr. Dobbs web site about this issue. You may want to search for it. [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Tobias Güntner Guest Posted: Wed Jul 21, 2004 7:49 pm    Post subject: Re: How to overload operator->* Maxim Yegorushkin wrote: Quote: For data members it returns a reference to the member. struct some { int i_; } s, *ps = &s; int some::*pi = &some::i_; int &ri = ps->*pi; Ah, of course... That's right. I totally forgot data members Thanks. Quote: But how to implement a generic operator->*? You can implement it using boost::function<> and boost::bind: template<class T> struct smart_ptr {/*...*/}; // for one arg member functions template<class T, class R, A0 inline function*(smart_ptr<T> const& p, R(T::*f)(A0)) { struct invoker { static R invoke(smart_ptr<T> const& p, R(T::*f)(A0), A0 const& a0) { return (p.get()->*f)(a0); } }; return function<R()>(bind(&invoker::invoke, p, f, _1)); } // add versions for N args, member functions, handle reference args Looks good. I'll have a closer look at boost::function<>. Thanks again :) Regards, Tobias [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. First time posters: Do this! ] Back to top Tobias Güntner Guest Posted: Wed Jul 21, 2004 7:50 pm    Post subject: Re: How to overload operator->* Joe wrote: Quote: Several years ago a read an article on the Dr. Dobbs web site about this issue. You may want to search for it. I do. I hope it's still online. Regards, Tobias [ See http://www.gotw.ca/resources/clcm.htm for info about ] [ comp.lang.c++.moderated. 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