this is the code:
NSNumber *taskId = [[self.taskList objectAtIndex:indexPath.row] valueForKey:#"identity"];
NSInteger *intTaskId = [[self.taskList objectAtIndex:indexPath.row] valueForKey:#"identity"];
self.taskList is an NSArray which filled with core data fetch request in ViewController's viewDidLoad method.
the taskId is: 1
the intTaskId is: 269303816
In actually, the value stored in core data is: 1
below is my questions:
1, I am confused why the NSInteger incorrect?
2, Should I have to replace NSInteger with NSNumber to avoid any other problems?
NSNumber is an object, whereas NSInteger is simply a typedef for a primitive (non-object) type (like int). NSInteger is not a subclass of NSNumber. Core Data returns numbers as instances of NSNumber. You're getting the weird NSInteger value because it's pointing to an object of type NSNumber but attempting to print it as if it were just an integer.
You'll need to replace NSInteger with NSNumber to avoid any problems. You could also use the intValue method on NSNumber to get back an NSInteger:
NSNumber *objTaskId = [[self.taskList objectAtIndex:indexPath.row] valueForKey:#"identity"];
NSInteger *intTaskId = [objTaskId intValue];
You'll need to do this if you want to do comparisons (greater than, equal too, smaller than) or arithmetic (you can't add an NSNumber to another NSNumber or an NSNumber to a primitive type like an int or float).
An other way please.
I have an array whole elements are doubles. How can I get the content out as double?e.g.
coordinate.latitude = [coords objectAtIndex:0];// value inside it is 61.2180556;
You can't store primitive doubles in NSArray. You have either stored them as NSNumber or NSString. In either case use:
coordinate.latitude = [[coords objectAtIndex:0] doubleValue];
I have the following construct, which I would like to simplify. I had to use a NSString (?) in order to get rid of the NSNumber vs. NSDecimalNumber compiler Warning.
NSDecimalNumber *ticksSinceSeventies = [NSDecimalNumber decimalNumberWithString:[NSString stringWithFormat:#"%d",[self timeIntervalSince1970]]];
Thanks for your Help!
If you are only using NSTimeIntervals, which are doubles, then the base 10 arithmetic that NSDecimalNumber affords you isn't really necessary. You can just use NSNumber instead.
Your constructor would therefore just be:
NSNumber *ticksSinceSeventies = [NSNumber numberWithDouble:[self timeIntervalSince1970]];
(assuming self is an NSDate subclass, or this is inside a category method).
Here is one way:
[NSDecimalNumber decimalNumberWithString:[[self timeIntervalSince1970] stringValue]];
There may be ways to simplify it still better, but i am not aware.
Is there a way to change the value contained in an NSNumber after it is created without making it point to a different NSNumber?
NSNumber *num = [NSNumber numberWithInt:0];
num = [NSNumber numberWithInt: 1]; // now num points to a different object, which I don't want. I want it the same object still, but different value.
NSNumber is immutable. Actually, it's a subclass of NSValue, and all NSValues are immutable.
No, you can't change the value of NSNumber.
See, for example, this post.
Now this must be easy, but how can sum two NSNumber? Is like:
[one floatValue] + [two floatValue]
or exist a better way?
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.
NSDecimalNumber (subclass of NSNumber) has all the goodies you are looking for:
– decimalNumberByAdding:
– decimalNumberBySubtracting:
– decimalNumberByMultiplyingBy:
– decimalNumberByDividingBy:
– decimalNumberByRaisingToPower:
...
If computing performance is of interest, then convert to C++ array std::vector or like.
Now I never use C-Arrays anymore; it is too easy to crash using a wrong index or pointer. And very tedious to pair every new [] with delete[].
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]);
The current top-voted answer is going to lead to hard-to-diagnose bugs and loss of precision due to the use of floats. If you're doing number operations on NSNumber values, you should convert to NSDecimalNumber first and perform operations with those objects instead.
From the documentation:
NSDecimalNumber, an immutable subclass of NSNumber, provides an object-oriented wrapper for doing base-10 arithmetic. An instance can represent any number that can be expressed as mantissa x 10^exponent where mantissa is a decimal integer up to 38 digits long, and exponent is an integer from –128 through 127.
Therefore, you should convert your NSNumber instances to NSDecimalNumbers by way of [NSNumber decimalValue], perform whatever arithmetic you want to, then assign back to an NSNumber when you're done.
In Objective-C:
NSDecimalNumber *a = [NSDecimalNumber decimalNumberWithDecimal:one.decimalValue]
NSDecimalNumber *b = [NSDecimalNumber decimalNumberWithDecimal:two.decimalValue]
NSNumber *result = [a decimalNumberByAdding:b]
In Swift 3:
let a = NSDecimalNumber(decimal: one.decimalValue)
let b = NSDecimalNumber(decimal: two.decimalValue)
let result: NSNumber = a.adding(b)
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]);
}
In Swift you can get this functionality by using the Bolt_Swift library https://github.com/williamFalcon/Bolt_Swift.
Example:
var num1 = NSNumber(integer: 20)
var num2 = NSNumber(integer: 25)
print(num1+num2) //prints 45