On 01/04/2019 12:46 AM, Alex wrote:
> On Friday, 4 January 2019 at 07:37:43 UTC, bauss wrote:
>> No, because you OVERRIDE A's foo().
>>
>> A does not exist. A is B and when you cast B to A you just tell the
>> compiler that the reference should only have A's signature available.
>>
>> You're not assigning B to A.
>
> Let's assume this is right. How to force a B object to behave like an A
> object?
Not possible by default. However, a B object is already behaving like an
A object because e.g. it overrides the foo() member function.
> I thought casting is a possible approach...
In this case, casting is using the B object through it's A interface.
The overridden behavior does not change. (Actually, that is possible in
languages that support multiple inheritance through multiple virtual
function pointer tables (vtbl) but D does not support that.)
Although I'm getting into implementation details here, I think it helps
with understanding the semantics. There is only one vtbl per class
object in D and the function entries are all filled in during
construction. So, a B object's 'foo' slot in that table is filled with
B.foo. So, such an object can foo() only as a B.
If there is such a need and B can indeed support behaving like an A, it
can do so itself by calling A.foo, not through vtbl, but directly:
class B : A {
override void foo() {
A.foo(); // Calling directly
}
}
By the way, do you have a use case in mind? Perhaps there are other ways
to achieve that.
Ali
