"Ary Borenszweig" wrote
> Christopher Wright escribió:
>> Steven Schveighoffer wrote:
>>> "Christopher Wright" wrote
>>>> Steven Schveighoffer wrote:
>>>>> Not exactly ;) The wrapped type is not equivalent to inheritance.
>>>> No, exactly:
>>>>
>>>> class Wrapper(T) : T
>>>> {
>>>> T opDot() {}
>>>> }
>>>
>>> class Wrapper(T) : T
>>> {
>>> }
>>>
>>> works just as good ;)
>>>
>>> -Steve
>>
>> True, but with opDot, you can swap out the real value. I can think of
>> cases in which this would be useful -- a sort of "I'll fill in this value
>> later" thing.
>
> I'm not sure that's the decorator pattern any more. In that pattern you
> implement or extend a class, and receive an instance of one in the
> constructor and forward all calls to that instance. If you do:
>
> class Wrapper(T) : T {
>
> private wrapped;
>
> this(T wrapped) {
> this.wrapped = wrapped;
> }
>
> T opDot() { return wrapped; }
>
> }
>
> that won't work because whenever you call a method of T on Wrapper(T),
> that will call Wrapper's method, since it has it, because it
> extends/implements T (opDot won't be triggered). If you remove
> inheritance, that will probably work, but you won't be able to make a
> Wrapper(T) behave like a T for the type system.
Exactly what I was going to say ;) If you inherit from a T, then return T
in an opDot, opDot will never be called unless you call opDot directly.
-Steve
