Re: derived class pointer to base class object
Rahul wrote:
On Dec 3, 9:27 pm, "Victor Bazarov" <v.Abaza...@comAcast.net> wrote:
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.
V
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Then, how do you suggest to type cast from Base class object to
derived class pointer?
I am not really sure how to tell you this... Have you been reading
what I wrote? You can only do it using 'dynamic_cast' or 'static_cast'.
However, in your case, since the object you originally create is NOT
of the derived class, 'dynamic_cast' will fail (return NULL) and
'static_cast' has undefined behaviour.
*I suggest* you _don't_ cast the pointer to the base class to
a pointer to derived class when there is *no reason for it*.
If you need an object of the derived class, create an object of that
type. Then you can convert its address to the pointer of the base
class and later 'static_cast' it back to the same derived class.
What is the problem you're trying to solve?
V
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