Reputation: 7804
I'm having a thread function which takes a weak_ptr<> and I pass my shared_ptr<> in the thread function.
Legally weak_ptr<> should not increment the reference count of shared_ptr<>, however, unless I typecast with weak_ptr<> while passing the same to the thread function, it increments the reference count (unexpected)
This behaviour happens only with thread functions and not with normal function calls.
Here is the code for thread function
void thrdfn(weak_ptr<int> wp) {
cout<<wp.use_count()<<endl; // Prints 2
}
int main() {
shared_ptr<int> sp = make_shared<int>();
thread th { thrdfn, (sp)};
th.join();
return 0;
}
However, when I typecast while creating thread, it behaves properly
void thrdfn(weak_ptr<int> wp) {
cout<<wp.use_count()<<endl; // Prints 1
}
int main() {
thread th { thrdfn, weak_ptr<int>(sp)}; // typecast
}
When I call the function as a normal function call, it works fine without typecasting
void thrdfn(weak_ptr<int> wp) {
cout<<wp.use_count()<<endl; // Prints 1
}
int main() {
shared_ptr<int> sp = make_shared<int>();
thrdfn(sp);
return 0;
}
Behaviour is same with multiple compilers
Upvotes: 3
Views: 276
Reputation: 302663
When you write construct a std::thread
(and I removed the superfluous parentheses):
thread th{thrdfn, sp};
What happens is:
The new thread of execution starts executing
std::invoke(decay_copy(std::forward<Function>(f)), decay_copy(std::forward<Args>(args))...);
where decay_copy is defined as
template <class T> std::decay_t<T> decay_copy(T&& v) { return std::forward<T>(v); }
Which is to say, your shared_ptr
is copied into the thread and you take a weak_ptr
off of that copy. So there are two shared_ptr
s: yours and the thread
's.
Upvotes: 2
Reputation: 5156
Avoid trying auto conversion from shared_ptr to weak_ptr, it may create extra shared_ptr. Below snippet should work fine :
shared_ptr<int> sp = make_shared<int>();
weak_ptr<int> weak = sp;
thread th { thrdfn, (weak)};
Upvotes: 0