成员函数可以通过哪些方式相互比较?

In what ways member functions may be compared to each other?(成员函数可以通过哪些方式相互比较?)

问题描述

我想知道是否可以将 2 个成员函数与<"进行比较操作员.我可以做=="，但在下面的情况下我不能使用它.我尝试将它们强制转换为 void*，但这也不起作用.

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<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?

推荐答案

如果你只是想任意将它们排序为 set/map 中的键，那么你可以reinterpret_cast 它们.您可能需要像 exact_int<sizeof(void (Foo::*bar)())>::type 这样的模板类，因为 指向成员函数的指针可以有有趣的大小.

If you just want to arbitrarily order them to be keys in a set/map, then you can reinterpret_cast them. You may need a template class like exact_int<sizeof(void (Foo::*bar)())>::type because pointers to member functions can have funny sizes.

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