这段代码安全吗？ 施放std :: vector <Derived*> 到std :: vector <Base*>

Question

Basically I've a class A and a class B : public A .

And I'd like to cast a std::shared_ptr<std::vector<A*> to a std::shared_ptr<std::vector<B*>

The problem is std::vector<B> doesn't inherit from std::vector<A> , and the smart_ptr neither. So I do a horrible cast :

  std::shared_ptr<VectorA> vector_a = * ((std::shared_ptr<VectorA>*)&vector_b);

The code compiles and runs there, but it is safe? http://liveworkspace.org/code/3dQTz1$0

Answer 1

This is a horrible solution. When you cast a vector of objects of one type to the other you can expect all kind of incorrect/undefined behaviour.

You should use vector of std::shared_ptr<A> from the beginning and initialize them with pointers to objects of type B there.

If it is not possible you can create a new vector holding std::weak_ptr<A> to avoid managing these objects twice.

Answer 2

Don't do that.

Consider that a vector of (pointers to) B is not a vector of (pointers to) A , or to be more precise, is not a universally valid substitution for a vector of (pointers to) A ; thus, there is a good reason why you cannot perform such a conversion.

Although it is true that all you have in a vector of B is indeed a set of objects which are (also( instances of A , consider the following algorithm:

void f(vector<A>& v)
{
    A a;
    v.push_back(a);
}

Now imagine you invoke f() with a vector of B . That would be an attempt to add an instance of a class that is not B to a collection which is supposed to contain only elements of type B .

The solution here is to make the code which accepts only a vector<A> flexible enough to work also on a vector<B> . In other words, you need to make it a template. For instance, your template could accept as arguments only vectors of a type which is derived from A . This is quite easy to enforce with some SFINAE techniques and type traits such as std::is_base_of .

Answer 3

Strictly speaking the dereferencing operations should succeed. Both vectors are pointer containers, so casting one to another, whilst unacceptable for production code, will still use the same dimensions and alignment.

C++ provides rich abstractions to avoid these shenanigans though, it would be better to populate the vector of derived objects as pointers to the base class.

Answer 4

You can still cast the elements:

A* a = vector_b->at (42);

gives you what you want.

Now, if you have written functions taking shared_ptr<vector<A*>> as arguments, then you have multiple solutions to take you out of this situation:

• use shared_ptr<A*[]> instead of shared_ptr<vector<A*>> (it works as you'd like to)
• take A** as arguments (and use vector_b->data () ) : it does not ties you to shared pointers.
• Better: take iterators which dereference as [smart] pointers to A (recall that operator-> chains)
• use std::vector<std::shared_ptr<A>> (wastes resources)