Question

I have a situation where I have the following, briefly:

class K {
  K clone() const{K cl; /* clone this to cl then */ return cl; }
};
K* call_clone()
{
  K k;
  return new K(k.clone());
}

after compiler optimization, will this be doubly copied?

Alternatively:
Would be an implementation with C*clone() more efficient?

Like this:

class K {
  K* clone() const { K*p=new K(); /* clone this to *p then */ return p; }
};
K* call_clone()
{
  K k;
  return k.clone();
}

(I ask this because even a shallow copy of K could be costly, it will be a class data struct in practice.)

Answer 1

after compiler optimization, will this be doubly copied?

It depends upon the compiler. Any modern compiler will not perform a double copy, this is called a Return Value Optimization. It does not depend on whether the definition of clone() is available or not.

Would be an implementation with C*clone() more efficient?

No, it won’t. Dynamic memory management (operators new/delete and alike) are very expensive. It might be even more expensive than performing a double-copy, depending on what your copy-constructor actually does.