如何创建固定大小的对象数组 [英] How to create a fixed-size array of objects

问题描述

在Swift中，我正在尝试创建一个包含64个SKSpriteNode的数组。我想先将它初始化为空，然后我将精灵放在前16个单元格中，最后16个单元格（模拟国际象棋游戏）。

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).

从我的理解在文档中，我会期望类似：

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

var sprites = SKSpriteNode（）[64];

或

var sprites4：SKSpriteNode [64];

但它不起作用。
在第二种情况下，我收到一条错误消息：尚不支持定长数组。这可能是真的吗？对我来说，这听起来像一个基本功能。
我需要通过索引直接访问元素。

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.

推荐答案

目前还不支持定长数组。这究竟意味着什么？并不是说你不能创建一个 n 的数组 - 显然你可以做让a = [1,2,3] 获取三个 Int 的数组。这意味着只是数组大小不能将声明为类型信息。

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.

如果你想要一个 nil s，你首先需要一个可选类型的数组 - [SKSpriteNode？] ，而不是 [SKSpriteNode ] - 如果声明一个非可选类型的变量，无论是数组还是单个值，它都不能是 nil 。 （还要注意 [SKSpriteNode？] 与 [SKSpriteNode]不同？ ...你想要一个选项数组，而不是一个可选的数组。）

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在设计上非常明确地要求对变量进行初始化，因为关于未初始化引用内容的假设是程序的一种方式在C（和其他一些语言）可以成为越野车。所以，你需要明确要求一个 [SKSpriteNode？] 数组，其中包含64 nil s：

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)

这实际上返回 [SKSpriteNode？]？，但是：可选的sprite数组。 （有点奇怪，因为 init（count：，repeatedValue：）不应该返回nil。）要使用数组，你需要打开它。有几种方法可以做到这一点，但在这种情况下，我更喜欢可选的绑定语法：

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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