Making an array of integers in iOS

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)

---

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]

---

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

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]);

If you want to use a NSArray, you need an Objective-C class to put in it - hence the NSNumber requirement.

That said, Obj-C is still C, so you can use regular C arrays and hold regular ints instead of NSNumbers if you need to.

You can use CFArray instead of NSArray. Here is an article explaining how.

CFMutableArrayRef ar = CFArrayCreateMutable(NULL, 0, NULL);
for (NSUInteger i = 0; i < 1000; i++)
{
  CFArrayAppendValue(ar, (void*)i);
}
CFRelease(ar); /* Releasing the array */

The same applies for the CoreFoundation version of the other containers too.