unique_ptr释放是否会导致内存泄漏？

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?

Answer 1

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.

Answer 2

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.