成员函数指针的行为含香函数指针,但你想调用他们时,除了参数之外,你还必须传递一个对象。
class Parrot
{
public:
void Eat()
{
cout << "Tsk, knick, tsk..." << endl;
}
void Speak()
{
cout << "On Captain, My Captain" << endl;
}
};
int main()
{
// Define a type: pointer to a member function of Parrot,taking no arguments and return void
typedef void(Parrot::*TpMemFun) ();
// Create an object of that type and initialize it with the address of Parrot::Eat
TpMemFun pActivity = &Parrot::Eat;
// Create a Parrot...
Parrot geronimo;
// a pointer to it...
Parrot *pGeronimo = &geronimo;
// Invoke the member function stored in Activity
// via an object. Notice the user of operator.*
(geronimo.*pActivity)();
// Same, via pointer. Now we use operator->*
(pGeronimo->*pActivity)();
// Change the activity
pActivity = &Parrot::Speak;
(geronimo.*pActivity)();
return 0;
}
你可以获取一般函数的 reference, 却无法获得成员函数的 reference。