做boost :: shared_ptr <T> 和boost :: shared_ptr <const T> 分享参考计数？

Question

There are several interesting questions on pitfalls with boost::shared_ptr s. In one of them, there is the useful tip to avoid pointing boost::shared_ptr<Base> and boost::shared_ptr<Derived> to the same object of type Derived since they use different reference counts and might destroy the object prematurely.

My question: Is it safe to have both boost::shared_ptr<T> and boost::shared_ptr<const T> point to the same object of type T , or will this cause the same problem?

Answer 1

It is perfectly safe.

The following code sample:

#include <iostream>
#include <boost/shared_ptr.hpp>

int main(int, char**)
{
  boost::shared_ptr<int> a(new int(5));
  boost::shared_ptr<const int> b = a;

  std::cout << "a: " << a.use_count() << std::endl;
  std::cout << "b: " << b.use_count() << std::endl;

  return EXIT_SUCCESS;
}

Compiles and run fine, and is perfectly correct. It outputs:

a: 2
b: 2

The two shared_ptr share the same reference counter.

---

Also:

#include <iostream>
#include <boost/shared_ptr.hpp>

class A {};
class B : public A {};

int main(int, char**)
{
    boost::shared_ptr<A> a(new B());
    boost::shared_ptr<B> b = boost::static_pointer_cast<B>(a);

    std::cout << "a: " << a.use_count() << std::endl;
    std::cout << "b: " << b.use_count() << std::endl;

    return EXIT_SUCCESS;
}

Behave the same way. You must, however never build your shared_ptr using a construct like this:

boost::shared_ptr<A> a(new B());
boost::shared_ptr<B> b(static_cast<B*>(a.get()));

a.get() gives the raw pointer and loses all information about reference counting. Doing this, you'll end up with two distinct (not linked) shared_ptr that use the same pointer but different reference counters.