Copied object changes attributes from the base class pointer it used to copy

Question

Consider the following classes:

 class Base {
 public:
     ... // include virtual destructor and rest of methods
     virtual void changeField(int val) = 0;
     virtual Base * clone() const = 0;
 };

 class Derived: public Base {
     int x;
 public:
     ... // include its destructor and rest of its methods
     void changeField(int val) { x = val; }
     Derived * clone() const { return new Derived(*this); }
 };

Suppose I have an existing Base * pointer bp that points to an Derived object. Then I call bp->clone() to make a copy and store the pointer of the resulting object in a Base * pointer, copyPointer.

When I try to changeField on the copyPointer, the value is changed, but the original object has its field also changed. Why is this? And what can I do to prevent this? Would I have to create an entirely new object from scratch?

Edit: Here is my main function in which I implement the described scenario

int main() {
     try {
        Base * copyPointer = bp->clone();
        copyPointer->changeField(5);
        cout << copyPointer->print() << endl;  //prints the field of Derived
        delete copyPointer;
      }
      catch (exception& e) {    // I also have an Exception class in my code
        cout << e.what() << endl;
      }
}

You don't show `changeField` or your copy constructors. What makes you think they are error free? — StoryTeller - Unslander Monica, Nov 13 '16 at 22:58
That example is not complete. What is `bp`? And there is nothing in that example which exhibits your concern. Anyway, look up "rule of three". — Peter, Nov 13 '16 at 22:59
What you posted is not a [mcve]. Post code that requires **no** further editing -- we should be able to take what you posted, copy it to a compiler's edit window, build the example, and see the error. — PaulMcKenzie, Nov 13 '16 at 23:00
If I haven't implemented copy constructors, wouldn't the default ones get called? — CSishard, Nov 13 '16 at 23:02
And what makes you think the defaults do the correct thing for your class? Do you hold any pointers? If you do, then the default copy constructor will happily have both objects point to the same thing. — StoryTeller - Unslander Monica, Nov 13 '16 at 23:03
I see, thank you for the clarification. I think I have figured out the cause of the behaviour I tried to describe. I'll try to present more clear and complete examples of code next time. — CSishard, Nov 13 '16 at 23:06

score 1 · Answer 1 · edited Nov 23 '16 at 21:16

Your assumption, that the function changeField() on the copyPointer changes the original object, is wrong!

I elaborated your example:

#include <iostream>
using std::cout;
using std::endl;

class Base {
     public:
     // include virtual destructor and rest of methods
     virtual void changeField(int val) = 0;
     virtual Base * clone() const = 0;
     virtual int print() const =0;
};

class Derived: public Base {
     int x;
     public:
     // include its destructor and rest of its methods
     Derived(int i):x(i)  {}
     void changeField(int val) { x = val; }
     Derived * clone() const { return new Derived(*this); }
     int print()const  { return x; }
 };
 int main() {
     Base* bp =new Derived(3);
     cout <<bp->print() <<endl;
     Base * copyPointer = bp->clone();
     copyPointer->changeField(5);
     cout <<copyPointer->print() <<endl;  //prints the field of Derived
     cout <<bp->print() <<endl;
 }

and the output is:

3
5
3

Worth to advise: use of keyword `override` in the derived class — Christophe, Nov 23 '16 at 21:34

Copied object changes attributes from the base class pointer it used to copy

1 Answers1