一个指向成员的指针的真实例子可以是std::shared_ptr的更窄的混叠构造函数:

template <typename T>
template <typename U>
shared_ptr<T>::shared_ptr(const shared_ptr<U>, T U::*member);

构造函数有什么用

假设你有一个结构体foo:

struct foo {
    int ival;
    float fval;
};

如果你给了一个foo对象一个shared_ptr对象，你可以使用构造函数将shared_ptr对象检索到它的成员ival或fval:

auto foo_shared = std::make_shared<foo>();
auto ival_shared = std::shared_ptr<int>(foo_shared, &foo::ival);

如果想将指针foo_shared->ival传递给某个需要shared_ptr的函数，这将非常有用

https://en.cppreference.com/w/cpp/memory/shared_ptr/shared_ptr

它是一个“指向成员的指针”——下面的代码说明了它的用法:

#include <iostream>
using namespace std;

class Car
{
    public:
    int speed;
};

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

    Car c1;
    c1.speed = 1;       // direct access
    cout << "speed is " << c1.speed << endl;
    c1.*pSpeed = 2;     // access via pointer to member
    cout << "speed is " << c1.speed << endl;
    return 0;
}

至于你为什么要这样做，它给了你另一种间接的层次，可以解决一些棘手的问题。但说实话，我从未在自己的代码中使用过它们。

编辑:我想不出一个令人信服的使用指针成员数据。指向成员函数的指针可以在可插拔的体系结构中使用，但是在这么小的空间里生成一个例子再次让我感到挫败。以下是我最好的(未经测试)尝试-一个Apply函数，在应用用户选择的成员函数到对象之前，会做一些前后处理:

void Apply( SomeClass * c, void (SomeClass::*func)() ) {
    // do hefty pre-call processing
    (c->*func)();  // call user specified function
    // do hefty post-call processing
}

c->*func周围的括号是必要的，因为->*操作符的优先级低于函数调用操作符。

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);
}

下面是一个例子，其中指向数据成员的指针可能很有用:

#include <iostream>
#include <list>
#include <string>

template <typename Container, typename T, typename DataPtr>
typename Container::value_type searchByDataMember (const Container& container, const T& t, DataPtr ptr) {
    for (const typename Container::value_type& x : container) {
        if (x->*ptr == t)
            return x;
    }
    return typename Container::value_type{};
}

struct Object {
    int ID, value;
    std::string name;
    Object (int i, int v, const std::string& n) : ID(i), value(v), name(n) {}
};

std::list<Object*> objects { new Object(5,6,"Sam"), new Object(11,7,"Mark"), new Object(9,12,"Rob"),
    new Object(2,11,"Tom"), new Object(15,16,"John") };

int main() {
    const Object* object = searchByDataMember (objects, 11, &Object::value);
    std::cout << object->name << '\n';  // Tom
}

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.

指向类的指针不是真正的指针;类是一个逻辑构造，在内存中没有物理存在，然而，当你构造一个指向类成员的指针时，它会给出一个指向该成员所在类的对象的偏移量;这给出了一个重要的结论:由于静态成员不与任何对象相关联，因此指向成员的指针不能指向静态成员(数据或函数) 考虑以下几点:

class x {
public:
    int val;
    x(int i) { val = i;}

    int get_val() { return val; }
    int d_val(int i) {return i+i; }
};

int main() {
    int (x::* data) = &x::val;               //pointer to data member
    int (x::* func)(int) = &x::d_val;        //pointer to function member

    x ob1(1), ob2(2);

    cout <<ob1.*data;
    cout <<ob2.*data;

    cout <<(ob1.*func)(ob1.*data);
    cout <<(ob2.*func)(ob2.*data);


    return 0;
}

来源:完整参考c++ - Herbert Schildt第四版