Re: Is this like calling a virtual function at construction time?
Sam <sakarab@yahoo.com> writes:
Hello!
I frequently read that at construction time "*this" has the type of the
class being constructed at the moment. This is usually written in
conjunction with calling virtual functions from constructors.
My case seams similar to the above but is not exactly the same. The
following code depicts my case (sort of, but the C++ mechanics
involved are the same).
#include <iostream> // for std::cout
class IPure
{
public:
virtual void Work() = 0;
virtual ~IPure() {} // empty
};
class Base : public IPure
{
private:
IPure *next;
public:
Base() : next(this) {} // empty
void CallNextWork() { next->Work(); }
virtual void Work() { std::cout << "Base"; }
};
class Derived : public Base
{
public:
virtual void Work() { std::cout << "Derived"; }
};
int main()
{
Base *base = new Derived();
base->CallNextWork();
delete base;
return 0;
}
In the above code, I want Derived::Work to be called. And so it happens
with two compilers I've tested it.
But I wonder. Since Base::next is initialized at "Base"s
construction time, when its type is "Base", why is it that it works
the way I want?
virtual function resolution depends on the type of the object *at the
time the function call is made*, not the type at the time the pointer
(if any) is initialized.
The virtual function call in question is made after the constructors
have finished and *base is fully constructed and of derived type. By
that time, the fact that *base was not yet of type Derived at the time
base->next was initialized to point to *base, is no longer important
Can I rely on this behavior? What does the C++ standard say about this?
[snip]
virtual functions are covered in section 10.3 .
--
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