How to increment a NSNumber

Question

How do I increment a NSNumber?

i.e. myNSNumber++

Answer 1

NSNumber objects are immutable; the best you can do is to grab the primitive value, increment it then wrap the result in its own NSNumber object:

NSNumber *bNumber = [NSNumber numberWithInt:[aNumber intValue] + 1];

Answer 2

Update: FYI, I personally like BoltClock's and DarkDusts's one-line answers better. They're more concise, and don't require additional variables.

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In order to increment an NSNumber, you're going to have to get its value, increment that, and store it in a new NSNumber.

For instance, for an NSNumber holding an integer:

NSNumber *number = [NSNumber numberWithInt:...];
int value = [number intValue];
number = [NSNumber numberWithInt:value + 1];

Or for an NSNumber holding a floating-point number:

NSNumber *number = [NSNumber numberWithDouble:...];
double value = [number doubleValue];
number = [NSNumber numberWithDouble:value + 1.0];

Answer 3

For anyone who is using the latest version of Xcode (writing this as of 4.4.1, SDK 5.1), with the use of object literals, you can clean the code up even a little bit more...

NSNumber *x = @(1);
x = @([x intValue] + 1);
// x = 2

Still kind of a pain to deal with the boxing and unboxing everything to do simple operations, but it's getting better, or at least shorter.

Answer 4

NSNumbers are immutable, you have to create a new instance.

// Work with 64 bit to support very large values
myNSNumber = [NSNumber numberWithLongLong:[myNSNumber longLongValue] + 1];

// EDIT: With modern syntax:
myNSNumber = @([myNSNumber longLongValue] + 1);

Answer 5

Put them all together, and you have:

myNSNumber = @(myNSNumber.intValue + 1);

Answer 6

I like the dot expression because it is more concise and that is why IOS supports it in the first place:

myNSNumber = [NSNumber numberWithInt:myNSNumber.intValue + 1];

Answer 7

If you're using it to store an integral value:

myNSNumber = @(myNSNumber.longLongValue + 1);

For floating point stuff the short answer is the following, but it's a bad idea to do this, you'll loose precision and if you're doing any kind of comparison later like [myNSNumber isEqual:@(4.5)] you might be surprised:

myNSNumber = @(myNSNumber.floatValue + 1);

If you need to do math on floating point numbers represented as objects in Objective-C (i.e. if you need to put them in arrays, dictionaries, etc.) you should use NSDecimalNumber.

Answer 8

Use a category to ease future use. Here is a basic idea.

 - (void)incrementIntBy:(int)ammount {
      self = [NSNumber numberWithInt:(self.intValue + ammount)];
 }

Answer 9

Lots of good info in the answers to this question. I used the selected answer in my code and it worked. Then I started reading the rest of the posts and simplified the code a lot. It’s a bit harder to read if you aren’t familiar with the changes in notation that were introduced in Xcode 5, but it’s a lot cleaner. I probably could make it just one line, but then it’s a little too hard to figure out what I’m doing.

I’m storing an NSNumber in a dictionary and I want to increment it. I get it, convert it to an int, increment it while converting it to an NSNumber, then put it back into the dictionary.

Old Code

 NSNumber *correct = [scoringDictionary objectForKey:@"correct"];
 int correctValue = [correct intValue];
 correct = [NSNumber numberWithInt:correctValue + 1];
 [scoringDictionary removeObjectForKey:@"correct"];
 scoringDictionary[@"correct"] = correct;

New code

NSNumber *correct = [scoringDictionary objectForKey:@"correct"];
scoringDictionary[@"correct"] = @(correct.intValue + 1);