No, you should be able to cast, as the method is found dynamically at run-time. Simple test program shows this.
I created a new project and edited the AppDelegate.h as here:
#import <Cocoa/Cocoa.h>
@interface SYNAppDelegate : NSObject <NSApplicationDelegate>
@property (assign) IBOutlet NSWindow *window;
@end
@interface ObjectA : NSObject {
}
- (NSInteger)doSomethingNew;
@end
@interface ObjectB : ObjectA
- (NSInteger)doSomethingOld;
@end
And edited the AppDelegate.m as here:
#import "SYNAppDelegate.h"
@implementation SYNAppDelegate
- (void)applicationDidFinishLaunching:(NSNotification *)aNotification
{
id obj = [[ObjectB alloc] init];
NSInteger i = [((ObjectB*) obj) doSomethingNew];
}
@end
@implementation ObjectA
- (NSInteger) doSomethingNew {
return 1;
}
@end
@implementation ObjectB
@end
Setting a breakpoint on the shows that variable i is set to 1 as expected, with
no runtime failure.
Paul
On Sep 24, 2013, at 8:52 AM, Kyle Sluder <k...@ksluder.com> wrote:
> On Sep 24, 2013, at 8:38 AM, Koen van der Drift <koenvanderdr...@gmail.com>
> wrote:
>
>> In my app I am using a 3rd party framework, and I have subclassed (objB) one
>> of the classes (objA) for additional functionality.
>>
>> At one point I am getting an NSCountedSet from the framework with objects
>> objA. When I enumerate these, I'd like to cast them as objB, to access the
>> additional functionality (doSomethingNew). But whatever I try, the objects
>> are always casted as objA.
>>
>> NSCountedSet *cs = [framework countSymbols];
>>
>> for (objB *obj in cs)
>> {
>> NSDictionary *ec = [objB doSomethingNew]; <<== throws error,
>> because objA doesn't know about doSomethingNew
>>
>> // etc
>> }
>>
>>
>> or:
>>
>> NSCountedSet *cs = [framework countSymbols];
>>
>> for (id *obj in cs)
>> {
>> NSDictionary *ec = [((ObjB*) obj) doSomethingNew]; <<== throws
>> error, because objA doesn't know about doSomethingNew
>>
>> // etc
>> }
>>
>>
>> Is it possible what I am trying to do?
>
> No, this doesn't make sense. Casting just tells the compiler "I know better
> than you and can guarantee you this expression is actually of this type". It
> doesn't "convert" objects from one type to another—how would it even do that?
>
> --Kyle Sluder
>
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