复制对象时切勿忘记copy基类的成员值

有基类Customer如下，

void logCall(const std::string& funcName);
class Customer {
public:
    //...
    Customer(const Customer& rhs);
    Customer& operator=(const Customer& rhs);
    //...

private:
    std::string name;
};

Customer::Customer(const Customer & rhs)
    : name(rhs.name)
{
    logCall("Customer copy constructor");
}

Customer& Customer::operator=(const Customer& rhs)
{
    logCall("Customer copy assignment operator");
    name = rhs.name;

    return *this;
}

倘若该类被PrioCustomer类继承，其构造如下，

class PrioCustomer: public Customer {
public:
    //...
    PrioCustomer(const PrioCustomer& rhs);
    PrioCustomer& operator=(const PrioCustomer& rhs);
    //...

private:
    int priority;
};

PrioCustomer::PrioCustomer(const PrioCustomer & rhs)
    : priority(rhs.priority);
{
    logCall("PrioCustomer copy constructor");
}

PrioCustomer& PrioCustomer::operator=(const PrioCustomer & rhs)
{
    logCall("PrioCustomer copy assignment operator");

    priority = rhs.priority;

    return *this;
}

初看好像该copy构造函数没问题，但这里需要指明的是PrioCustomer类只复制了其自身定义的成员变量，并没有copy基类Customer的成员变量，
其基类Customer的成员变量只是以默认参数初始化之。

正确的构造应该，

PrioCustomer::PrioCustomer(const PrioCustomer & rhs)
    : Customer(rhs),
       priority(rhs.priority);
{
    logCall("PrioCustomer copy constructor");
}

PrioCustomer& PrioCustomer::operator=(const PrioCustomer & rhs)
{
    logCall("PrioCustomer copy assignment operator");

    Customer::operator=(rhs);
    priority = rhs.priority;

    return *this;
}