Mind Twister......

Hi,
  I got this question from one of the user group.. I found it quite interesting.... So I thought you people must give it a try.... I guess i know the answer but i am not that confident about it...... Its just a puzzle to puzzle you.....

class base
{
public:

      virtual void func1 ()
      {
            cout << "Base: func1" << endl;
      }
      
      void func2 ()
      {
            cout << "Base: func2" << endl;
      }
};

class derived
{
public:
      void func1 ()
      {
            cout << "Derived: func1" << endl;
      }
      
      virtual void func2 ()
      {
            cout << "Derived: func2" << endl;
      }
};


int main(void)
{
      base* pb1;
      derived d1;

      pb1=(base*)&d1;

      pb1->func1 ();
      pb1->func2 ();

 return 0 ;
}

Guess the output first..... And then try the explaination part......

Don't think the answer but know the answer ... ;-)

Dennis
LVL 5
dennis_georgeAsked:
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AxterCommented:
The result of this code is undefined according to the C++ standard.

The base class is not a base class of the derived class.
So forcing the compiler to use the wrong class type pointer for your object leads to undefined behavior.
0

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DanRollinsCommented:
I agree with Axter.  As shown, you might as well call strcpy(), with the two parameters being a pointer to a function and a pointer to a void pointer... something might happen, but whatever it is, would be meaningless.

Perhaps the question is mis-stated.  If the object named 'derived'  is actually derived from the object named 'base' then the question might have some meaning.

-- Dan
0
gwapCommented:
Like axter said, the result is undefined according to the standard, however, due to the most common implementation of the C++ standard, your output should be :
Derived:func2
Base:func2

I don't need to explain the Base:func2 outcome as it is merely a static link.
The Derived:func2 outcome is because, for class base, the pb1->func2 call is translated to "the first function in the virtual function pointer table."

This virtual function pointer table is the table of Derived; furthermore, the first (and only) entry in the table of Derived is that of virtual function Derived::func2.

But like Axter said, this is implementation defined.
0
havman56Commented:
dennis,

Axter said is absoluetly correct behaviour undefined.

kindly dont try to test experts  knowledge in EE.
see section 10.3 of C++ standard

0
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