|
|
Posted by Csaba Gabor on 11/15/07 14:01
Steve, thanks for your awesome exposition on VB interfaces, replete
with example. This type of post is all too rare on the web.
Csaba
On Nov 14, 8:10 pm, "Steve" <no....@example.com> wrote:
> "Alex" <d_key...@hotmail.com> wrote in message
....
> a difference, yes. something that will make your code quit bombing and still
> get the Factory functionality...probably not. all an interface (IFactory in
> this case) is, is a contract...a definition of what *all usable* objects
> that *implement* that interface will provide a caller, i.e. your php code.
> make sense? IFactory will provide you no other functionality than that.
> here's an example in vb.
>
> public interface iFactory
> public property foo as integer
> end interface
>
> public class sawMill
> implements iFactory
> public property get foo() implements iFactory.foo
> get
> return 666
> end get
> set (byval value as integer)
> # do nothing...however, set is *required*
> # because iFactory is read/write
> end property
> public function saw()
> debug.print "sawing..."
> end function
> end class
>
> public class brewery
> implements iFactory
> private myFoo as integer
> public property get foo() implements iFactory.foo
> get
> return myFoo
> end get
> set (byval value as integer)
> myFoo = value
> end property
> public function brew()
> debug.print "brewing " & myFoo
> end function
> end class
>
> in the above, both completely unrelated classes (sawMill and brewery) have
> iFactory.foo as part of their definition. how they implement them is
> different, but each must have a read and write foo.
>
> if i simply say:
>
> private myFactory as iFactory
>
> and then try:
>
> myFactory.foo()
>
> it will do nothing. myFactory is an interface that has no working parts.
> however, if i do:
>
> private myFactory as iFactory = new brewery
> myFactory.foo = 15
>
> i'll get somewhere. notice that the above works whether or not i set
> myFactory to new brewery OR sawMill. both have a foo interface. however, if
> i continue the above code with:
>
> myFactory.saw()
>
> it'll blow up. both saw and brew are specific interfaces defined by each
> class respectively...not by iFactory. let's say that myFactory is set
> somewhere unbeknownst to me, how would i take specific action?
>
> public function factoryToString(byval factory as iFactory, byval fooValue as
> integer)
> factory.foo = fooValue
> if typeof factory is sawMill then
> factory.saw()
> end if
> if typeof factory is brewery then
> factory.brew()
> end if
> end function
>
> anyway...i think i got off track, but does that help you know that you're
> going to have to get at an actual Factory object in the list...which doesn't
> solve your memory problem?
Navigation:
[Reply to this message]
|