Re: Full code example (Re: Returning derived class pointer from base class pointer without casts)

From:

JiiPee <no@notvalid.com>

Newsgroups:

comp.lang.c++

Date:

Sun, 08 Mar 2015 13:23:54 +0000

Message-ID:

<P3YKw.997936$4b6.54375@fx44.am4>

On 08/03/2015 13:16, Paavo Helde wrote:

JiiPee <no@notvalid.com> wrote in news:7FXKw.298809$SK1.210976@fx17.am4:

but how would you implement the visitor (the idea)? I tried:

class AnimalBase
{
..
virtual void accept(Farm* farm) = 0;
}
;;;;
template <typename T>
void Dog<T>::accept(Farm* farm)
{
farm->visit(this);
}

then :
farm.getAnimal(0)->accept(&farm);

But how does this help to return the Dog? I dont understand how will
that return Dog. Is this what you meant? Where would the dog be
returned?

If you want derived class (Dog) returned at compile time you must be used
a compile-time feature, i.e. a template. In other words, the function
returning Dog must be a template, and it must be instantiated with the
Dog type, again at compile-time.

I see. So something like:
if(farm.animal[0].type == DOG)
Dog* dog = farm.getAnimal<Dog>[0];

so must ask it via template argument.

For example, the Dog class itself knows it is Dog, at compile time, so if
it there is a virtual function of Dog which is calling some template
function, it can easily instantiate it as Dog. I guess this might be what
?? Tiib meant by using the visitor pattern.

hth
Paavo