Getting derived classes

[Sean Kelly]

dsimcha wrote:
> == Quote from Andrei Alexandrescu (SeeWebsiteForEmail at erdani.org)'s article
>> dsimcha wrote:
>>> I know that, in std.traits, there's a template that spits out all base classes
>>> for a given class.  Based on reading the source, it seems to work based on a
>>> rather interesting use case for is expressions.  Is there any equivalent way
>>> to do the opposite:  For any given position in a class hierarchy, to get a
>>> tuple of all possible descendants?
>> That's not possible in general because in D the derived classes form an
>> open set. I guess it could be done at runtime via reflection (not
>> implemented afaik), but not at compile time.
>> Andrei
> 
> Not quite sure I understand why this has to be the case.  Somewhere in the docs,
> Walter explicitly says that, since D knows about the whole class tree, the
> compiler can figure out which functions in a hierarchy can be made non-virtual.
> If this is the case, why is the compiler not able to know about the whole class
> tree for purposes of creating lists of derived classes?

I think the actual implementation of this is somewhat different.  Rather 
than relying on knowledge of the entire class tree, the compiler simply 
has to know what type a particular function is being called on.  For 
example:

class C
{
     void fn() {}
}

void main()
{
     auto c = new C;
     c.fn();
}

Here, the compiler knows that it's dealing with an instance of C rather 
than a possibly unknown derived class so it can call fn() directly 
rather than going through the vtbl.  However:

void callFn( C c )
{
     c.fn();
}

Here, the compiler doesn't know what the underlying type is so it must 
go through the vtbl.


Sean