在 C++ 中快速将基对象的所有成员分配给派生对象

Question

假设我们有一个基础 class 和一个衍生 class:

class Base {
  string s1;
  string s2;
  ...
  string s100; // Hundreds of members
};

class Derived : public Base{
  string s101;
};

我想将 Base 对象 base 分配给 Derived 对象 derived。我知道我们不能只使用运算符“=”将基对象分配给它的派生对象。我的问题是：我们是否必须将所有成员逐一复制？喜欢：

derived.s1 = base.s1;
derived.s2 = base.s2;
...
derived.s100 = base.s100;

有没有更快或更简洁的方法来做到这一点？重载 operator= 与返回的基础对象？

Answer 1

I know we can't just use operator "=" to assign a base object to its derived object.

当然可以（在这个问题的上下文中）：

static_cast<Base &>(derived)=base;

库存示例：

class Base {};

class Derived : public Base {};

void foo()
{
    Derived d;
    Base b;

    static_cast<Base &>(d)=b;
}

Answer 2

I know we can't just use operator "=" to assign a base object to its derived object

这不是真的。

Do we have to make copies of all the members one by one? Like: base.s1 = derived.s1; base.s2 = derived.s2; ... base.s100 = derived.s100;

不是真的。正如 Danh 在第一条评论中提到的。

base = derived

就足够了，因为它执行隐式动态向上转换（即从派生指针转换为基指针）。参见 http://www.cplusplus.com/doc/tutorial/typecasting/

Answer 3

I want to assign a Base object base to a Derived object derived.

为其提供重载operator=：

class Derived : public Base {
    Derived& operator=(const Base& b) { 
        Base::operator=(b); // call operator= of Base
        s101 = something;   // set sth to s101 if necessary
        return *this; 
    }
};

那你可以

Base b;
// ...
Derived d;
// ...
d = b;

在 C++ 中快速将基对象的所有成员分配给派生对象

Quickly assign all the members of a base object to a derived object in C++

c++

inheritance

assignment-operator