shared_ptr中的dynamic_cast？

Question

I have two classes A and B, B inherits from A.

If I have a shared_ptr<A> object which I know is really a B subtype, how can I perform a dynamic cast to access the API of B (bearing in mind my object is shared_ptr, not just A?

Answer 1

If you just want to call a function from B you can use one of these:

std::shared_ptr<A> ap = ...;
dynamic_cast<B&>(*ap).b_function();
if (B* bp = dynamic_cast<B*>(ap.get()) {
    ...
}

In case you actually want to get a std::shared_ptr<B> from the std::shared_ptr<A> , you can use use

std::shared_ptr<B> bp = std::dynamic_pointer_cast<B>(ap);

Answer 2

use dynamic_pointer_cast

example copied from above link

// static_pointer_cast example
#include <iostream>
#include <memory>

struct A {
  static const char* static_type;
  const char* dynamic_type;
  A() { dynamic_type = static_type; }
};
struct B: A {
  static const char* static_type;
  B() { dynamic_type = static_type; }
};

const char* A::static_type = "class A";
const char* B::static_type = "class B";

int main () {
  std::shared_ptr<A> foo;
  std::shared_ptr<B> bar;

  bar = std::make_shared<B>();

  foo = std::dynamic_pointer_cast<A>(bar);

  std::cout << "foo's static  type: " << foo->static_type << '\n';
  std::cout << "foo's dynamic type: " << foo->dynamic_type << '\n';
  std::cout << "bar's static  type: " << bar->static_type << '\n';
  std::cout << "bar's dynamic type: " << bar->dynamic_type << '\n';

  return 0;
}

output

foo's static  type: class A
foo's dynamic type: class B
bar's static  type: class B
bar's dynamic type: class B

Answer 3

可能最好的方法是使用标准函数来转换shared_ptr