Does unique_ptr release cause memory leaks?

Question

I'm confused about unique_ptr.release().

My goal is to cast a unique_ptr of a base class to a unique_ptr of a derived class.

So I found this question and the answer is

Derived *tmp = dynamic_cast<Derived*>(basePointer.get());
std::unique_ptr<Derived> derivedPointer;
if(tmp != nullptr)
{
    basePointer.release();
    derivedPointer.reset(tmp);
}

or

std::unique_ptr<Derived>
    derivedPointer(static_cast<Derived*>(basePointer.release()));

Then I was wondering what happen to the base pointer after basePointer.release();.

Based on this question, I understand that it causes a memory leak.

Am I right?

score 12 · Accepted Answer · answered Dec 08 '16 at 18:11

12

Am I right?

No.

Calling release() doesn't leak anything, it just signals that you are taking control of it.

If you leak a pointer after explicitly releasing it from a smart pointer, that's your fault.

answered Dec 08 '16 at 18:11

Useless

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So if I cast the pointer to another unique_ptr, it should be fine then? – Marc Dec 08 '16 at 18:12
Yep, in your code it behaves exactly as Benjamin indicates - I just went for the general answer rather than the specific. – Useless Dec 08 '16 at 18:13
No problem at all :) – Useless Dec 08 '16 at 18:16

score 6 · Answer 2 · answered Dec 08 '16 at 18:11

6

A memory leak happens when you lose track of the last pointer to a dynamically allocated object before you delete it. Since you copied the pointer to tmp first, you didn't lose track of it when you called release(). So no, there is no memory leak here.

answered Dec 08 '16 at 18:11

Benjamin Lindley

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Thank you very much! Exactly what I wanted to make sure. – Marc Dec 08 '16 at 18:13

Does unique_ptr release cause memory leaks?

2 Answers2