将基对象返回到向上转换的派生类的最合适方法是什么？
struct B {}; // B contains data members
struct D : B {}; // D doesn't contain ANY data member
B g_b; // global object
D& fun () // want to return by reference ONLY
{
return <???>(g_b); // how ???
}
[注意：我想避免重载构造函数（或赋值），例如 D(const B&)。]
struct B {}; // B contains data members
struct D : B {}; // D doesn't contain ANY data member
B g_b; // global object
D& fun () // want to return by reference ONLY
{
return <???>(g_b); // how ???
}
[Note: I want to avoid overloading constructor (or assignment) such as D(const B&).]
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你试图做的事情是非法的。
g_b不是D。What you're trying to do is illegal.
g_bis not aD.没有合适的演员。这实际上是未定义的行为。
有关详细信息，请参阅此主题：向下转型基本类型
注意：术语是向下转型 当你将基类转换为派生类时；当将派生类强制转换为基类时，将使用术语向上转换。
No suitable cast. That is in fact undefined behavior.
For detail, see this topic: Downcasting a base type
Note : the term is downcast when you cast base to derived class; and the term upcast is used when you cast derived to base class.
这是未定义的行为。
您可以使用
dynamic_cast执行基类指针/引用到派生类指针/引用的安全向下转换。如果是指针，则返回 null；如果是引用，则抛出异常。That is undefined behavior.
You can use
dynamic_castfor performing safe down casting of Base class pointer/reference to derived class pointer/reference. It returns a null in case of pointers or throws an exception in case of references.