派生类如何从基类继承静态函数?

2022-01-05 00:00:00 inheritance static c++ 
struct TimerEvent
{
   event Event;
   timeval TimeOut;
   static void HandleTimer(int Fd, short Event, void *Arg);
};

HandleTimer 需要是静态的，因为我将它传递给 C 库 (libevent).

HandleTimer needs to be static since I'm passing it to C library (libevent).

我想继承这个类.这怎么办?

I want to inherit from this class. How can this be done?

谢谢.

推荐答案

您可以轻松继承该类:

class Derived: public TimerEvent {
    ...
};

但是，您不能在子类中覆盖 HandleTimer 并期望它起作用:

However, you can't override HandleTimer in your subclass and expect this to work:

TimerEvent *e = new Derived();
e->HandleTimer();

这是因为静态方法在 vtable 中没有条目，因此不能是虚拟的.但是，您可以使用void* Arg"将指针传递给您的实例……例如:

This is because static methods don't have an entry in the vtable, and can't thus be virtual. You can however use the "void* Arg" to pass a pointer to your instance... something like:

struct TimerEvent {
    virtual void handle(int fd, short event) = 0;

    static void HandleTimer(int fd, short event, void *arg) {
        ((TimerEvent *) arg)->handle(fd, event);
    }
};

class Derived: public TimerEvent {
    virtual void handle(int fd, short event) {
        // whatever
    }
};

这样，HandleTimer 仍然可以在 C 函数中使用，只需确保始终将真实"对象作为void* Arg"传递.

This way, HandleTimer can still be used from C functions, just make sure to always pass the "real" object as the "void* Arg".

相关文章