Dear Eljay,
thank you for your reply!
To fix your example, do this:
B y = B();
You are right. This creates a B on the stack and B::f() gets called.
Nevertheless, here virtual functions are not involved. The vtable is not
used when calling y.f()
My example created an A on the stack, and assigned a B to it, without
turning the A into a B. Right!
But, I wanted to use virtual functions with class objects purley living
on the stack _without_ doing some sort of casting:
B y = B();
dynamic_cast<A*>(&y)->f(); // B::f() called
(*(A*)(&y)).f(); // B::f() called
Nobody would want to write this. But somebody might want to use
objects with virtual functions and define hers object on the stack, like:
vector<A> ve;
ve.push_back(A());
ve.push_back(B());
ve.push_back(A());
iterate over it and call ::f(). Then always A::f() gets called, because
its a vector of A. But a B is also an A. Thatfor I can insert it into the
vector.
This example can only be realized with
vector<A*>.
I think, its not possible to make use of the vtable with objects
living on the stack.
--
Frank
