Reputation: 11901

Question regarding C++ Lists

list dog; ............. ............

So I added many dog objects to it. If I call dog.pop_front();

Does memory automatically gets deallocated ? For the object that I popped out ?

So If I call

list<Dog*> dog2;
dog2.push_back(dog.front());

and then I will call dog.pop_front() So this will work? I will assume Dog as type struct.

Upvotes: 0

Answers (7)

fredoverflow

Reputation: 263128

I suggest a list<shared_ptr<Dog> > which takes care of deleting the dogs.

Upvotes: 2

R Samuel Klatchko

Reputation: 76541

You keep asking about this sequence:

list<Dog*> dog2;
dog2.push_back(dog.front());  // time 1
dog.pop_front();              // time 2

At time1, both dog2 and dog have a pointer to the same object.

At time2, the pointer to that object is removed from dog and is only in dog2.

Assuming you originally created that object with new Dog, the object will not be freed until you explicitly free it by calling delete ptr

Upvotes: 1

Thomas

Reputation: 181745

The memory for the Dog object does not get deleted; this you'll have to do yourself.

However, the memory for the pointer, of type Dog*, as well as any "list node" object wrapped around it, will be automatically deleted by the list class.

Upvotes: 2

Naveen

Reputation: 73443

When you insert a Dog* into the list a copy of the pointer is created and insrted into the list. When you pop from the list memory allocated to this pointer is released. The memory allocated for the Dog object is not released as it was not allocated by the list. You have to release it yourself.

Upvotes: 0