Implementation of lock()

Question

According to boost:

To access the object, a weak_ptr can be converted to a shared_ptr using the shared_ptr constructor or the member function lock.

Again, from boost:

shared_ptr<T> lock() const;

//Returns: expired()? shared_ptr<T>(): shared_ptr<T>(*this).

As far as I understand, returning shared_ptr<T>(*this) means creating a new shared_ptr with reference count of 1; And this is definitely not what we want. So probably I'm not understand it right. Will someone explain? Thanks!

Answer 1

No, that is actually the point of shared_ptr - the copied instance will point to the same underlying data and increment the reference count for both instances.

That means that shared_ptr<T>(*this) will create an additional shared_ptr instance pointing to the same data and will increment the reference count for both this and the new instance.

---

It is actually more complicated in the real code, as the original shared_ptr data is accessed through the weak_ptr instance, but effectively the original shared_ptr data is shared at the end (with the shared reference count increased in all existing copies of the particular shared_ptr object).