How to create a fixed-size array of objects

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In Swift, I am trying to create an array of 64 SKSpriteNode. I want first to initialize it empty, then I would put Sprites in the first 16 cells, and the last 16 cells (simulating an chess game).

From what I understood in the doc, I would have expect something like:

var sprites = SKSpriteNode()[64];

or

var sprites4 : SKSpriteNode[64];

But it doesn't work. In the second case, I get an error saying: "Fixed-length arrays are not yet supported". Can that be real? To me that sounds like a basic feature. I need to access the element directly by their index.

Fixed-length arrays are not yet supported. What does that actually mean? Not that you can't create an array of n many things — obviously you can just do let a = [ 1, 2, 3 ] to get an array of three Ints. It means simply that array size is not something that you can declare as type information.

If you want an array of nils, you'll first need an array of an optional type — [SKSpriteNode?], not [SKSpriteNode] — if you declare a variable of non-optional type, whether it's an array or a single value, it cannot be nil. (Also note that [SKSpriteNode?] is different from [SKSpriteNode]?... you want an array of optionals, not an optional array.)

Swift is very explicit by design about requiring that variables be initialized, because assumptions about the content of uninitialized references are one of the ways that programs in C (and some other languages) can become buggy. So, you need to explicitly ask for an [SKSpriteNode?] array that contains 64 nils:

var sprites = [SKSpriteNode?](repeating: nil, count: 64)

This actually returns a [SKSpriteNode?]?, though: an optional array of optional sprites. (A bit odd, since init(count:,repeatedValue:) shouldn't be able to return nil.) To work with the array, you'll need to unwrap it. There's a few ways to do that, but in this case I'd favor optional binding syntax:

if var sprites = [SKSpriteNode?](repeating: nil, count: 64){
    sprites[0] = pawnSprite
}

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The best you are going to be able to do for now is create an array with an initial count repeating nil:

var sprites = [SKSpriteNode?](count: 64, repeatedValue: nil)

You can then fill in whatever values you want.


In Swift 3.0 :

var sprites = [SKSpriteNode?](repeating: nil, count: 64)

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Declare an empty SKSpriteNode, so there won't be needing for unwraping

var sprites = [SKSpriteNode](count: 64, repeatedValue: SKSpriteNode())

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This question has already been answered, but for some extra information at the time of Swift 4:

In case of performance, you should reserve memory for the array, in case of dynamically creating it, such as adding elements with Array.append().

var array = [SKSpriteNode]()
array.reserveCapacity(64)

for _ in 0..<64 {
    array.append(SKSpriteNode())
}

If you know the minimum amount of elements you'll add to it, but not the maximum amount, you should rather use array.reserveCapacity(minimumCapacity: 64).

Fixed-length Array, Fixed-length Array. old array. Once an array object has been constructed, the number of cells does not change. You must construct an array with a length that will accommodate all the data Will the following statement make the array larger? When your array is a collection of string or integers (value types), sometimes you will want to update the values in the array as you loop over them. Most of the loops above use a variable in the loop that holds a copy of the value. If you update that variable, the original value in the array is not updated.

For now, semantically closest one would be a tuple with fixed number of elements.

typealias buffer = (
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode,
    SKSpriteNode, SKSpriteNode, SKSpriteNode, SKSpriteNode)

But this is (1) very uncomfortable to use and (2) memory layout is undefined. (at least unknown to me)

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Comments
  • Thanks, I did try that one, but I had forget the "?". However, I am still unable to change the value? I tried both: 1) sprites[0] = spritePawn and 2) sprites.insert(spritePawn, atIndex:0).
  • Surprise! Cmd-click sprites in your editor/playground to see its inferred type — it's actually SKSpriteNode?[]?: an optional array of optional sprites. You can't subscript an optional, so you have to unwrap it... see edited answer.
  • That's quite odd indeed. As you mentioned, I don't think the array should be optional, as we explicitly defined it as ?[] and not ?[]?. Kind of annoying having to unwrap it every time I need it. In any case, this seem to work: var sprites = SKSpriteNode?[](count: 64, repeatedValue: nil); if var unwrappedSprite = sprites { unwrappedSprite[0] = spritePawn; }
  • Syntax has changed for Swift 3 and 4, please see other answers below
  • is there any way to declare an array of the fixed size?
  • @AlexanderSupertramp no, there is no way to declare a size for an array
  • @アレックス There is no way to declare a fixed size for an array, but you can certainly create your own struct that wraps an array that enforces a fixed size.
  • Be careful with this. It will fill the array with the same instance of that object (one might expect distinct instances)
  • Ok, but it solves the OP question, also, knowing the array is filled with the same instance object then you will have to deal with it, no offense.
  • The problem with this is it's a "let" array so it's technically constant and that doesn't compile. You will have to create a "var" array first and then initialize a constant array from the contents of the "var" array.
  • @iseletsky Yeah, I saw. I'll fix it. It was still just extra information. You can assign the var to a let afterwards.
  • How is that better than the array? To me it's a hack that doesn't even solve the problem: you could very well do a tasks[65] = foo in both this case and the case of an array from the question.