Re: derived class pointer to base class object

From:

Pete Becker <pete@versatilecoding.com>

Newsgroups:

comp.lang.c++

Date:

Mon, 3 Dec 2007 11:33:17 -0500

Message-ID:

<2007120311331775249-pete@versatilecodingcom>

On 2007-12-03 11:27:57 -0500, "Victor Bazarov" <v.Abazarov@comAcast.net> said:

Rahul wrote:

I was just playing around virtual functions and landed up with the
following,

class Base1
{
public: virtual void sample()
{
printf("base::sample\n");
};
};

class Derived1: public Base1
{
public: void sample()
{
printf("derived::sample\n");
};
};

int main()
{
Derived1 *ptr = reinterpret_cast<Derived1 *> (new Base1());
ptr->sample();
return(0);
}

this invokes the sample() of base version, but if i declare sample as
a ordinay (non-virtual) function, the sample() of derived class is
invoked. I'm wondering how? Can anyone help in this regard?

The code 'ptr->sample()' has undefined behaviour because casting from
a base class to a derived class is NOT what 'reinterpret_cast' is for.
Either 'static_cast' or 'dynamic_cast' are used for that.

In your case 'dynamic_cast' should return NULL since the object is
NOT of type 'derived', which should suggest that 'static_cast' is not
what you need either. You only use 'static_cast' in this situation
if you *know exactly* that the pointer was obtained by converting
some derived class object into the base.

And just to complete the circle: replacing reinterpret_cast with
static_cast in the original code still results in undefined behavior.

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)