Why has NSNumber such strange retainCounts?

Question

NSNumber* n = [[NSNumber alloc] initWithInt:100];
NSNumber* n1 = n;

In the code above, why is the value of n\'s retainCount set to 2? In the second

Answer 1

Based on this link here, it's possible that there's some optimization going on under the covers for common NSNumbers (which may not happen in all implementations hence a possible reason why @dizy's retainCount is 1).

Basically, because NSNumbers are non-mutable, the underlying code is free to give you a second copy of the same number which would explain why the retain count is two.

What is the address of n and n1? I suspect they're the same.

NSNumber* n = [[NSNumber alloc] initWithInt:100];

NSLog(@"Count of   n : %i",[n retainCount]);

NSNumber* n1 = n;

NSLog(@"Count of   n : %i",[n retainCount]);
NSLog(@"Count of   n1: %i",[n1 retainCount]);
NSLog(@"Address of n : %p", n);
NSLog(@"Address of n1: %p", n1);

Based on your update, that link I gave you is almost certainly the issue. Someone ran a test and found out that the NSNumbers from 0 to 12 will give you duplicates of those already created (they may in fact be created by the framework even before a user requests them). Others above 12 seemed to give a retain count of 1. Quoting:

From the little bit of examination I've been able to do, it looks as if you will get "shared" versions of integer NSNumbers for values in the range [0-12]. Anything larger than 12 gets you a unique instance even if the values are equal. Why twelve? No clue. I don't even know if that's a hard number or circumstantial.

Try it with 11, 12 and 13 - I think you'll find 13 is the first to give you a non-shared NSNumber.