How to add two NSNumber objects?

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Now this must be easy, but how can sum two NSNumber? Is like:

[one floatValue] + [two floatValue]

or exist a better way?

6 Answers

There is not really a better way, but you really should not be doing this if you can avoid it. NSNumber exists as a wrapper to scalar numbers so you can store them in collections and pass them polymorphically with other NSObjects. They are not really used to store numbers in actual math. If you do math on them it is much slower than performing the operation on just the scalars, which is probably why there are no convenience methods for it.

For example:

NSNumber *sum = [NSNumber numberWithFloat:([one floatValue] + [two floatValue])];

Is blowing at a minimum 21 instructions on message dispatches, and however much code the methods take to unbox the and rebox the values (probably a few hundred) to do 1 instruction worth of math.

So if you need to store numbers in dicts use an NSNumber, if you need to pass something that might be a number or string into a function use an NSNumber, but if you just want to do math stick with scalar C types.

You can use

NSNumber *sum = @([first integerValue] + [second integerValue]);

Edit: As observed by ohho, this example is for adding up two NSNumber instances that hold integer values. If you want to add up two NSNumber's that hold floating-point values, you should do the following:

NSNumber *sum = @([first floatValue] + [second floatValue]);

Why not use NSxEpression?

NSNumber *x = @(4.5), *y = @(-2);

NSExpression *ex = [NSExpression expressionWithFormat:@"(%@ + %@)", x, y];
NSNumber *result = [ex expressionValueWithObject:nil context:nil];

NSLog(@"%@",result); // will print out "2.5"

You can also build an NSExpression that can be reused to evaluate with different arguments, like this:

NSExpression *expr = [NSExpression expressionWithFormat: @"(X+Y)"];
NSDictionary *parameters = [NSDictionary dictionaryWithObjectsAndKeys:x, @"X", y, @"Y", nil];
NSLog(@"%@", [expr expressionValueWithObject:parameters context:nil]);

For instance, we can loop evaluating the same parsed expression, each time with a different "Y" value:

 for (float f=20; f<30; f+=2.0) {
    NSDictionary *parameters = [NSDictionary dictionaryWithObjectsAndKeys:x, @"X", @(f), @"Y", nil];
    NSLog(@"%@", [expr expressionValueWithObject:parameters context:nil]);
 }
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