如何返回转换为NSInteger的NSNumber

Question

With NSNumber being a class, I am trying to convert it to an NSInteger to do some computations. In NSLog, it shows that I am converting and doing the multiplication correct. However, when I got to return doubler as a regular NSInteger, I get "Implicit conversion of 'NSInteger' (aka 'long') to 'NSNumber* ' is disallowed with ARC". Where am I going wrong and what do I do to make this correct?

- (NSNumber *) numberThatIsTwiceAsBigAsNumber:(NSNumber *)number {

NSInteger doubler = [number integerValue] * 2;
NSLog(@"%ld", (long)doubler);

return doubler;
}

EDIT: For those curious, this is how I solved it:

NSInteger unboxing = [number integerValue] * 2;
NSNumber *boxing = [NSNumber numberWithInteger:unboxing];

return boxing;

Answer 1

You need to return the number as NSNumber, not NSInteger. So, convert the NSInteger to NSNumber before returning.

return @(doubler);

Answer 2

Change your return type to

- (NSInteger)

If you intend to continue to use it as such, or explicitly cast it.