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.
