你以后可以在任何实例上访问这个成员:

int main()
{    
  int Car::*pSpeed = &Car::speed;    
  Car myCar;
  Car yourCar;

  int mySpeed = myCar.*pSpeed;
  int yourSpeed = yourCar.*pSpeed;

  assert(mySpeed > yourSpeed); // ;-)

  return 0;
}

请注意，您确实需要一个实例来调用它，因此它不像委托那样工作。 它很少被使用，我这么多年来可能用过一两次。

通常使用接口(即c++中的纯基类)是更好的设计选择。

你以后可以在任何实例上访问这个成员:

int main()
{    
  int Car::*pSpeed = &Car::speed;    
  Car myCar;
  Car yourCar;

  int mySpeed = myCar.*pSpeed;
  int yourSpeed = yourCar.*pSpeed;

  assert(mySpeed > yourSpeed); // ;-)

  return 0;
}

请注意，您确实需要一个实例来调用它，因此它不像委托那样工作。 它很少被使用，我这么多年来可能用过一两次。

通常使用接口(即c++中的纯基类)是更好的设计选择。

我认为，只有当成员数据相当大(例如，另一个相当庞大的类的对象)，并且您有一些外部例程，只对该类的对象引用起作用时，才会想要这样做。你不想复制成员对象，所以这让你可以传递它。

这是我能想到的最简单的例子，它传达了这个特性很少相关的情况:

#include <iostream>

class bowl {
public:
    int apples;
    int oranges;
};

int count_fruit(bowl * begin, bowl * end, int bowl::*fruit)
{
    int count = 0;
    for (bowl * iterator = begin; iterator != end; ++ iterator)
        count += iterator->*fruit;
    return count;
}

int main()
{
    bowl bowls[2] = {
        { 1, 2 },
        { 3, 5 }
    };
    std::cout << "I have " << count_fruit(bowls, bowls + 2, & bowl::apples) << " apples\n";
    std::cout << "I have " << count_fruit(bowls, bowls + 2, & bowl::oranges) << " oranges\n";
    return 0;
}

这里需要注意的是传递给count_fruit的指针。这样就不必单独编写count_apples和count_oranges函数。

IBM有更多关于如何使用它的文档。简单地说，您使用指针作为类的偏移量。你不能在它们所指向的类之外使用这些指针，所以:

  int Car::*pSpeed = &Car::speed;
  Car mycar;
  mycar.*pSpeed = 65;

It seems a little obscure, but one possible application is if you're trying to write code for deserializing generic data into many different object types, and your code needs to handle object types that it knows absolutely nothing about (for example, your code is in a library, and the objects into which you deserialize were created by a user of your library). The member pointers give you a generic, semi-legible way of referring to the individual data member offsets, without having to resort to typeless void * tricks the way you might for C structs.

另一个应用是侵入式列表。元素类型可以告诉列表它的next/prev指针是什么。所以列表不使用硬编码的名称，但仍然可以使用现有的指针:

// say this is some existing structure. And we want to use
// a list. We can tell it that the next pointer
// is apple::next.
struct apple {
    int data;
    apple * next;
};

// simple example of a minimal intrusive list. Could specify the
// member pointer as template argument too, if we wanted:
// template<typename E, E *E::*next_ptr>
template<typename E>
struct List {
    List(E *E::*next_ptr):head(0), next_ptr(next_ptr) { }

    void add(E &e) {
        // access its next pointer by the member pointer
        e.*next_ptr = head;
        head = &e;
    }

    E * head;
    E *E::*next_ptr;
};

int main() {
    List<apple> lst(&apple::next);

    apple a;
    lst.add(a);
}