How to increment a NSNumber

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);

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];

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.

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];