Hello!
Interested in a question.
Object "A" inherited object "B". And you need to return the object link "B". Is this possible in this language? Here is an example:

>class B
>{
>    protected int a;
>    public void step() {};
>}
>
>class A : B
>{
>    public override step()
>    {
>        import std.random;
>        a = uniform(5,100);
>    }
>}
>
>void main()
>{
>     B* b;
>     A  a;
>     a.step();
>     b = a.B; // @FixMe: Here you need a link to object B from object A.
>}

Sorry to trouble you, thanks in advance!

On Thursday, 6 February 2020 at 12:15:17 UTC, Mihail Lorenko wrote:
>>     B* b;

A pointer to a class is a rare thing in B since they are already automatic references. Just use `B b;` and ten `b = a` will just work.

On Thursday, 6 February 2020 at 12:37:01 UTC, Adam D. Ruppe wrote:
> On Thursday, 6 February 2020 at 12:15:17 UTC, Mihail Lorenko wrote:
>>>     B* b;
>
> A pointer to a class is a rare thing in B since they are already automatic references. Just use `B b;` and ten `b = a` will just work.

Thanks for the answer!
Well, apparently I have to get around my problem. Thanks again!

On 2/6/20 8:05 AM, Mihail Lorenko wrote:
> On Thursday, 6 February 2020 at 12:37:01 UTC, Adam D. Ruppe wrote:
>> On Thursday, 6 February 2020 at 12:15:17 UTC, Mihail Lorenko wrote:
>>>>     B* b;
>>
>> A pointer to a class is a rare thing in B since they are already automatic references. Just use `B b;` and ten `b = a` will just work.
> 
> Thanks for the answer!
> Well, apparently I have to get around my problem. Thanks again!

Not sure, but you might misunderstand. B is already a pointer in D (like Java, D's class instances are always references).

And derived classes implicitly cast to base classes. So the a.B syntax is not necessary.

To illustrate further:

A a = new A; // note, instances by default are null, you need to new them
B b = a; // implicit cast

assert(a is b); // they are the same object
b.step(); // virtual call, calls B.step

assert(a.a != 0); // note, protected allows access from same module.

-Steve

On 2/6/20 10:11 AM, Steven Schveighoffer wrote
> b.step(); // virtual call, calls B.step

I meant, calls A.step;

-Steve