shared_ptr中的dynamic_cast？

Question

我有两个A和B类,B继承自A.

如果我有一个shared_ptr<A>我知道的对象实际上是B子类型,我怎样才能执行动态转换来访问B的API(请记住我的对象是shared_ptr,而不仅仅是A？

Answer 1

如果您只想从中调用函数,B可以使用以下方法之一:

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

如果你真的想获得一个std::shared_ptr<B>从std::shared_ptr<A>,你可以使用使用

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

Answer 2

使用 dynamic_pointer_cast

从上面链接复制的示例

// 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;
}

产量

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