I have the following and it throws me an incompatible error integer to pointer ...
int *v = [numero intValue];
NSLog(@"Numero %d",v);
Where number is a NSNumber .
How would you do to print a int ? And a NSInteger ?
I have the following and it throws me an incompatible error integer to pointer ...
int *v = [numero intValue];
NSLog(@"Numero %d",v);
Where number is a NSNumber .
How would you do to print a int ? And a NSInteger ?
I understand that numero is declared as NSNumber with something of this style:
@property (strong, nonatomic) NSNumber *numero;
And what you want is to assign it to i which is a int .
The problem is that NSNumber is an object and the variable i is a int that is a primitive type. This means that numero will be a pointer to some place in the memory, that's why the * in the declaration, but i is not an object, so you do not need the * in the declaration.
If you do this, it will work properly for you:
int i = [self.numero integerValue];
NSLog(@"numero %d", i);