c++ - 成员函数可以通过哪些方式相互比较？

我想知道是否可以将2个成员函数与“ void * ，但这也不起作用。

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<Receiver, Sender>*)_other;
        if (receiver < other->receiver) {
            return true;
        } else if (receiver == other->receiver && function < other->function) {
            return true; // this line gives the error
        }
        return false;
    }
};

有什么想法吗？

最佳答案

如果您只想随意将它们作为集合/映射中的键，则可以对其进行reinterpret_cast。您可能需要像exact_int<sizeof(void (Foo::*bar)())>::type这样的模板类，因为pointers to member functions can have funny sizes。

关于c++ - 成员函数可以通过哪些方式相互比较？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/6662917/