Reputation: 5352
If you take a lock on std::weak_ptr
:
class foo {};
auto s = std::make_shared<foo>();
std::weak_ptr<foo> w{ s };
s = nullptr;
s.reset(new foo());
auto l = w.lock(); // try to get previous foo
std::cout << "locked: " << (l != nullptr) << std::endl;
Output:
locked: 0
Once lock returns nullptr
, is there ever a condition under which it could return non-null, or is it effectively dead?
My test program suggests once the originally allocated object's reference count is zero then no, weak_ptr
will always return a nullptr.
Upvotes: 10
Views: 2693
Reputation: 219345
#include <iostream>
#include <memory>
int
main()
{
using namespace std;
shared_ptr<int> p(new int{3});
weak_ptr<int> w = p;
p = nullptr;
auto l = w.lock();
if (l == nullptr)
cout << "l = nullptr\n";
else
cout << "l = " << *l << '\n';
p.reset(new int{4});
w = p;
l = w.lock();
if (l == nullptr)
cout << "l = nullptr\n";
else
cout << "l = " << *l << '\n';
}
Portable output:
l = nullptr
l = 4
So yes, if the weak_ptr
is assigned from a non-empty shared_ptr
, it can become "reanimated from the dead."
Maybe that isn't what you meant. But it is what you asked.
Upvotes: 5
Reputation: 39390
is there every a condition under which it could return non-null, or is it effectively dead?
Well, it's "almost" dead at that point. lock
calls expired
in a way equivalent to this:
expired() ? shared_ptr<T>() : shared_ptr<T>(*this)
Then, looking at the expiration condition:
Equivalent to
use_count() == 0
. The destructor for the managed object may not yet have been called, but this object's destruction is imminent (or may have already happened).
Either way, there's no possibility for that specific weak_ptr
to be reassigned to another value.
Upvotes: 9