In what ways member functions may be compared to each other?

Question

I would like to know if I can compare 2 member functions with the "<" operator. I can do "==" but I can't use it in the case below. I tried casting them to void* but that won't work either. template <class Receiver, class Sender> class CallBack2 : public ICallBack2 { protected: Receiver* receiver; void(Receiver::*function)(Sender*); Sender* sender; public: CallBack2(Receiver* _receiver, void(Receiver::*_function)(Sender*), Sender* _sender) : receiver(_receiver), function(_function), sender(_sender) {}; virtual ~CallBack2() {}; virtual void callBack() { (receiver->*function)(sender); } virtual bool operator<(const ICallBack2* _other) const { CallBack2<Receiver, Sender>* other = (CallBack2*)_other; if (receiver < other->receiver) { return true; } else if (receiver == other->receiver && function < other->function) { return true; // this line gives the error } return false; } }; Any ideas please? JavaScript questions and answers, JavaScript questions pdf, JavaScript question bank, JavaScript questions and answers pdf, mcq on JavaScript pdf, JavaScript questions and solutions, JavaScript mcq Test , Interview JavaScript questions, JavaScript Questions for Interview, JavaScript MCQ (Multiple Choice Questions)

Answer 1

C++03 § 5.9, which covers the semantics of the built-in <, >, <= and >=, doesn't mention pointers to members and states: Other pointer comparisons are unspecified. According to § 8.3.3, 3 The type "pointer to member" is distinct from the type "pointer", As a result, we can conclude the result of relational operators applied to pointers to members (whether functions or fields) is unspecified. Note that "unspecified behavior" is different from "undefined behavior", but still means you can't usefully apply the operators as different implementations may have different results. "Unspecified" basically means the implementation gets to define the behavior.