Copy Constructors and Inheritance

[1yR4jE8b]

I have a bit of a conundrum: If have a base class with a properly defined (deep-)copy constructor, and I derive from that class and the only new member variable is a single pointer, and if I WANT the pointer to simply be a shallow copy of that pointer when it is copy-constructed...do I need to implement the copy constructor of the derived class? (Example below)

class Bar
{
public:
     Bar(const Bar& other);  //this is implemented
private:
     //stuff here, some of it is also dynamically allocated memory
};

class Foo : public Bar
{
     //Is the compiler's default copy-constructor for this class sufficient?
private:
     int* ptr; //this is not deep copied upon class copy-construction
};

[kempofighter]

Can you use a debugger? It'd take about 5 minutes to write a small program, compile it, and debug it. Set a break point or add a cout to your base class's copy constructor and see what happens. It seems like you wouldn't have to but I'm not sure. I could be wrong. I am not clear on how you are going to properly destroy the object though. That could be tricky. if it is just a shared pointer in the derived class then you just have to make sure that somehow your program can shut down cleanly. Anyway I wouldn't wait around for an answer since you can test this in about 5 minutes.

[1yR4jE8b]

I have tried not implementing the copy constructor, in the derived class and the default copy-ctor seems to work fine. I was just wondering about the semantics of a copy constructor, I thought Copy Constructors weren't supposed to automatically call copy constructors of base classes which is why I was asking.

I guess it's only automatic if the compiler gives it to you.

[jsmith]

Correct.

If you provide your own copy constructor, then you have to ensure to call the base class copy
constructor otherwise the default one will be called.

[1yR4jE8b]

sweet, now I know for sure. That's what I thought happened, I just wanted to make sure by asking.

Thanks, guys