在C ++中，是否允许将其强制转换为仅添加非虚拟方法的派生类？

Question

I'd like to add non-virtual methods to an existing class A that I can't change. Therefore I create a class B inherited from A and add the methods I want. Now, if I have an object of type A , can I just regard it as an object of type B ? Say, is the following code legal?:

class A { <...> };

class B: public A {
    void f();
  };

A a();
void g( const B& );
void h() { g( static_cast<B&&>(a()) ); }

It does compile and work, however I was wondering if it's guaranteed by the standard to work as expected. I don't see why not, yet it doesn't feel very clean..

Answer 1

That is undefined behavior. From [expr.static.cast] :

An lvalue of type “cv1 B”, where B is a class type, can be cast to type “reference to cv2 D”, where D is a class derived (Clause 10) from B, if a valid standard conversion from “pointer to D” to “pointer to B” exists (4.10), cv2 is the same cv-qualification as, or greater cv-qualification than, cv1, and B is neither a virtual base class of D nor a base class of a virtual base class of D. The result has type “cv2 D”. An xvalue of type “cv1 B” may be cast to type “rvalue reference to cv2 D” with the same constraints as for an lvalue of type “cv1 B”. If the object of type “cv1 B” is actually a subobject of an object of type D, the result refers to the enclosing object of type D. Otherwise, the behavior is undefined.

a isn't actually a suboject of an object of type B - it's really just of type A , so the behavior is undefined.

What's to stop you from simply adding a non-member function:

void f(A& ); 

If the answer is that you need to access private or protected members of A , then there was intent to make those members unaccessible to you!