Reputation: 14918
I understand that the shared_ptr
class automatically manages dynamic objects.
Here, I have a function f
that returns a const shared_ptr<int>
to an int 2
.
I have two versions of main
that differ in one place. Version A saves the return value of f
into a shared pointer, while version B saves into shared pointer reference.
using namespace std;
const std::shared_ptr<int> f() {
const std::shared_ptr<int> ret = std::make_shared<int>(2);
return ret;
}
int main () {
const std::shared_ptr<int> p = f(); // A
// const std::shared_ptr<int> &p = f(); // B
cout << p.use_count() << *p << endl; // prints 1 2
return 0;
}
Both versions print 1 2
. I am okay with version A because p
is the last shared_ptr
pointing at the int
, therefore use_count
is 1
.
Question: Why is use_count
equal to 1
for version B? Where is the last existing shared_ptr
?
Upvotes: 0
Views: 244
Reputation: 275385
In C++ if you assign a temporary directly into a &&
or const&
, reference lifetime extension kicks in and the temporary lasts as long as reference does.
This only works when your reference is a local variable1.
1 Or in certain cases in certain compilers where you use aggregate construction of struct
s. But don't use that, because it is one of the spots where a "transparent" forwarding constructor behaves fundamentally different than an aggregate, so otherwise innocuous changes later can break your code.
Upvotes: 1