Making an array of integers in iOS

Question

If you want to make an array of integers, can you use NSInteger? Do you have to use NSNumber? If so, then why?

Answer 1

You can use a plain old C array:

NSInteger myIntegers[40];  for (NSInteger i = 0; i < 40; i++)     myIntegers[i] = i;  // to get one of them NSLog (@"The 4th integer is: %d", myIntegers[3]); 

Or, you can use an NSArray or NSMutableArray, but here you will need to wrap up each integer inside an NSNumber instance (because NSArray objects are designed to hold class instances).

NSMutableArray *myIntegers = [NSMutableArray array];  for (NSInteger i = 0; i < 40; i++)     [myIntegers addObject:[NSNumber numberWithInteger:i]];  // to get one of them NSLog (@"The 4th integer is: %@", [myIntegers objectAtIndex:3]);  // or NSLog (@"The 4th integer is: %d", [[myIntegers objectAtIndex:3] integerValue]); 

Answer 2

C array:

NSInteger array[6] = {1, 2, 3, 4, 5, 6}; 

Objective-C Array:

NSArray *array = @[@1, @2, @3, @4, @5, @6]; // numeric values must in that case be wrapped into NSNumbers 

Swift Array:

var array = [1, 2, 3, 4, 5, 6] 

This is correct too:

var array = Array(1...10) 

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NB: arrays are strongly typed in Swift; in that case, the compiler infers from the content that the array is an array of integers. You could use this explicit-type syntax, too:

var array: [Int] = [1, 2, 3, 4, 5, 6] 

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If you wanted an array of Doubles, you would use :

var array = [1.0, 2.0, 3.0, 4.0, 5.0, 6.0] // implicit type-inference 

or:

var array: [Double] = [1, 2, 3, 4, 5, 6] // explicit type