考虑这段代码。我已经为Person定义了一个复制构造函数，它将复制除了一个数据之外的所有数据，即它的DontCopy数据成员。但是我似乎不能用赋值运算符完成同样的事情。除了下面我已经做过的之外，还有什么最好的方法来完成它呢？目前，我只手动设置了name成员。如果Person有数百个成员需要复制，只有一个成员不需要复制，该怎么办？复制构造函数会处理这个问题，而赋值运算符不会。

#include <iostream>
#define show(variable) std::cout << #variable << " = " << variable << '\n';

struct DontCopy {
    class Person& subject;
    DontCopy (Person& p) : subject(p) { }
};

class PersonDataDoCopy {
protected:
    std::string name;
    PersonDataDoCopy (const std::string& n) : name(n) { }
    PersonDataDoCopy() = default;
};

class Person : private PersonDataDoCopy {
    DontCopy dontCopy{*this};
public:
    Person (const std::string& n) : PersonDataDoCopy(n) { }
    Person (const Person& other) : PersonDataDoCopy(other) { }  // 'dontCopy' is not copied from 'other'.
    Person (Person&& other) : PersonDataDoCopy(std::move(other)) { }  // 'dontCopy' is not copied from 'other'.
    Person& operator= (const Person& other) {
        if (this == &other)
            return *this;
//      static_cast<PersonDataDoCopy>(*this) = PersonDataDoCopy(other);  // Does not copy data from PersonDataDoCopy
//      *this = static_cast<Person&>(PersonDataDoCopy(other));  // Does not compile.
        name = other.name;  // Need more general solution than this.
        return *this;
    }
    Person() = default;
    void display() const { show(name)  show(this)  show(&dontCopy.subject); }
};

int main() {
    Person bob("Bob");
    bob.display();
    std::cout << '\n';
    Person clone1(bob);
    clone1.display();
    std::cout << '\n';
    Person clone2("");
    clone2 = bob;
    clone2.display();
}

输出：

name = Bob
this = 0x6ffde0
&dontCopy.subject = 0x6ffde0

name = Bob
this = 0x6ffdd0
&dontCopy.subject = 0x6ffdd0

name = Bob
this = 0x6ffdc0
&dontCopy.subject = 0x6ffdc0