CODEWITHSUNDEEP
c++pointersc++11atomic
pauluss86
pauluss86

Reputation: 464

Pointer ownership with atomic variables

Does an atomic variable, 'containing' a pointer, take ownership of the pointer?

Consider the following snippet:

{
    std::atomic<Foo*> bar(new Foo());
}

// `bar' went out of scope, did it delete pointer to instance of Foo?

Of course I could derive and delete it myself, or work around it in another way; but that's not the point.

What is the defined behavior here, if any?

Upvotes: 1

Views: 1125

Answers (2)

Jonathan Wakely
Jonathan Wakely

Reputation: 171263

No. std::atomic<T*> has a trivial destructor that does nothing, so it could not delete anything if it owned it.

std::atomic<int> doesn't "own" the integer, it just stores a value, and similarly std::atomic<int*> just stores a value, with no ownership or freeing implied.

Upvotes: 2

Caesar
Caesar

Reputation: 9841

No, the only thing std::atomic guarantees is that the object will be free from data races. So you will have free the memory that the pointer points to your self.

If you want a managed dynamic memory container, then either use a unique_ptr or shared_ptr.

Upvotes: 5

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