c++smart-pointersauto-ptr

What is the difference between *ptr and *ptr.get() when using auto_ptr?


Why would I use get() with *, instead of just calling *?

Consider the following code:

auto_ptr<int> p (new int);
*p = 100;
cout << "p points to " << *p << '\n';           //100

auto_ptr<int> p (new int);
*p.get() = 100;
cout << "p points to " << *p.get() << '\n'; //100

Result is exactly the same. Is get() more secure?


Solution

  • Practically no difference.

    In case of *p, the overloaded operator* (defined by auto_ptr) is invoked which returns the reference to the underlying object (after dereferencing it — which is done by the member function). In the latter case, however, p.get() returns the underlying pointer which you dereference yourself.

    I hope that answers your question. Now I'd advise you to avoid using std::auto_ptr, as it is badly designed — it has even been deprecated, in preference to other smart pointers such as std::unique_ptr and std::shared_ptr (along with std::weak_ptr).