On Jun 11, 2008, at 12:01 AM, John Engelhart wrote:
This whole business about '(NSArray *) means NSArray AND any of it's
subclasses' is the result of sloppy thinking and confusing 'able to'
with 'as per spec'.
No, you are wrong. It really is "per spec." That is an intentional
capability in both Objective-C and C++: You can return an instance of
a class, or any of its subclasses, wherever an instance of that class
is specified precisely because classes should be designed to be
substitutable in this fashion. This is the essence of polymorphism in
object-oriented programming and is summarized by the Liskov
substitution principle.
Languages like Smalltalk and Ruby take this even further with "duck
typing," where they lack type declaration entirely and an object's
interface is defined entirely by its expected behavior.
If you declare a method prototype as '-(NSArray *)resultsArray',
then you have explicitly communicated that a NSArray is going to be
returned. Not a NSMutableArray. Not 'Jimmies groovy array with red
pin stripes'. A NSArray. Period. A NSMutableArray != a NSArray.
If you're going to be returning (or accepting) more than a single
class, you use id, which clearly communicates your intentions.
This is absolutely, categorically incorrect.
The Objective-C language, for the purposes of type checking, treats
assignment, passing and return of subclasses as equivalent to the
assignment, passing and return of superclasses.
It does so specifically so instances of a subclass may be assigned,
passed and returned wherever an instance of a superclass is required.
A return type of NSArray * means something very explicit. "But all
you can expect is an object that behaves like a NSArray, so
returning a subclass of NSArray, like NSMutableArray, is perfectly
legal!" You're exactly right that all I can expect is an object
that behave like a NSArray. /Exactly/ like a NSArray. Not sorta.
Not almost. Exactly. Because that's what you explicitly stated.
You are incorrect. It means all you can expect is an object that
obeys NSArray's API contract. Note that it has been pointed out, at
LEAST once in this thread, that you will NEVER receive an actual
NSArray because NSArray is actually a class cluster.
You keep insisting that Objective-C is designed in and behaves in a
way in which it simply does not. Please stop.
-- Chris
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