If this is what your want to do I think you would be
better off with
lass Base {
...
virtual
..
}
class D1 : public Base {
}
Base* bp1 = new D1(..some
thing..);
Base* bp2;
ar << bp1;
ar >> bp2;
this might require registration or export
though.
Robert Ramey
Robert
Ramey wrote:
> Oh, that should be easy:
>
> D1
*d1p;
>
> ar << d1;
> ar >>
*d1p;
>
> Robert Ramey
Hi,
At
> ar <<
d1;
> ar >> bp; //
but "bp" points already to an object of type D1
I meant : "bp" is "Base* bp"
So, if I call ar
>> *bp; the Base::load will be called, not the derived
D1:load.
Actually more precisely, I intend to do
so
lass Base {
...
virtual ..
}
class D1 : public Base {
}
Base* bp1 = new D1(..some thing..);
Base* bp2 = new D1(..some thing else..);
ar << *bp1;
ar >> *bp2;
This means for all my
serializable classes I have a kind of "=" operator
defined.
regards,
Andrei
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