如何投射shared_ptr <a> to shared_ptr<b> where B is derived from A?

Question

I have a std::list container, holding shared pointers of say class A. I have another class, say B, that is derived from A.

I currently have code that does this to populate the container:

shared_ptr<B> b = shared_ptr<B>(new B);
container.push_back(b)

This works fine.

Question is, how can I retrieve the shared_ptr < B > that was initially pushed back to the container?

Doing the following

   list<shared_ptr<A> >::iter anIter = myContainer.begin();
   shared_ptr<B> aB = *(anIter);

do not compile. I get error

Cannot convert 'A * const' to 'B *'

Any tips?

Answer 1

如果您知道自己已经检索到了B，正如您的问题所示，那么您可以使用 constructor ：

dynamic_cast<>()

如果有疑问，您可以使用 static_pointer_cast<>() 。但与传统的 container.push_back(make_shared<A>()); for (auto i = container.begin(); i!=container.end(); i++) { shared_ptr<B> spb = dynamic_pointer_cast<B>(*i); if (spb) spb->showb(); // at least one virtual function else cout << "the pointer was not to a B"; } 一样，只有当您的类具有多态时才会起作用，即您至少有一个虚函数：

static_cast<>()

对于普通指针，原则类似于dynamic_cast<>()和const_pointer_cast<>()，请参阅dynamic_pointer_cast<>()。顺便说一句，对于普通指针，还有一个const_cast<>()，就像 view.backgroundColor = UIColor.colorWithPatternImage(UIImage(named: "background.jpg")) 一样。

这是dynamic_cast and static_cast in C++

Answer 2

您可以使用std::dynamic_pointer_cast。

你这样使用它：

std::shared_ptr<Base> basePtr;
std::shared_ptr<Derived> derivedPtr = std::dynamic_pointer_cast<Derived>(basePtr);