Re: Why is the constructor of B called?

[Artur Skawina via Digitalmars-d-learn]

On 09/24/15 13:26, Marc Schütz via Digitalmars-d-learn wrote:
> On Thursday, 24 September 2015 at 01:01:09 UTC, Nicholas Wilson wrote:
>> On Wednesday, 23 September 2015 at 21:25:15 UTC, tcak wrote:
>>> On Wednesday, 23 September 2015 at 21:14:17 UTC, Adam D. Ruppe wrote:
>>>> On Wednesday, 23 September 2015 at 21:08:37 UTC, tcak wrote:
>>>>> I wouldn't expect B's constructor to be called at all unless "super" is 
>>>>> used there.
>>>>
>>>> "If no call to constructors via this or super appear in a constructor, and 
>>>> the base class has a constructor, a call to super() is inserted at the 
>>>> beginning of the constructor. "
>>>>
>>>>
>>>> from http://dlang.org/class.html#constructors
>>>>
>>>> the idea is to make sure the base class construction work is done too.
>>>
>>> Is there any way to prevent this behaviour?
>>>
>>> Quickly checked whether Java acts in the same way. Answer is yes.
>>
>> You might be able to swap out the vtbl entry  for a stub call it and trick 
>> the compiler and swap it back, but...
> 
> Urgh...
> 
> If you can modify the base class, and you really need it, you can check the 
> dynamic type:
> 
> class Base {
>     this() {
>         if(!cast(Base) this) return;
>         // do the initialization
>     }
> }

If you're going to do this then you can just use overloading.
That will both avoid the runtime check and require the hack
to be explicitly enabled in the derived class. IOW:

   class B {
       this() {
           writeln("B.constructor");
           foo();
       }

       struct SkipBCtor {}
       this(SkipBCtor) {}

       void foo() {
           writeln("B.foo");
       }
   }

   class D : B {
       this() {
           super(SkipBCtor());
           writeln("D.constructor");
       }

       override void foo() {
           writeln("D.foo overrides B.foo");
       }
   }

artur