成员函数在什么方面可以相互比较? [英] In what ways member functions may be compared to each other?
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问题描述
我想知道如果我可以比较2个成员函数与<运算符。我可以做==但我不能使用它在下面的情况。我尝试将它们转换为 void * ,但不起作用。
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;
}
};
有任何想法吗?
推荐答案
如果你只是想任意地将它们设置为set / map中的键,那么你可以 reinterpret_cast
。您可能需要一个类似 exact_int< sizeof(void(Foo :: * bar)())> :: type
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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