Question

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.

Answer 1

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