ios 返回数组的 swift 函数
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/31268209/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
swift function returning an Array
提问by pete
Im learning Swift and I can understand how to create a simple function that takes in an Array and returns an Array. Heres my code:
我正在学习 Swift，我可以理解如何创建一个简单的函数，它接受一个数组并返回一个数组。这是我的代码：
func myArrayFunc(inputArray:Array) -> Array{
var newArray = inputArray
// do stuff with newArray
return newArray
}
The red error I get is: Reference to generic type 'Array" requires arguments in <>
我得到的红色错误是：对泛型类型“数组”的引用需要 <> 中的参数
回答by Kirsteins
In Swift Arrayis generic type, so you have to specify what type array contains. For example:
在 Swift 中Array是泛型类型，所以你必须指定数组包含什么类型。例如：
func myArrayFunc(inputArray:Array<Int>) -> Array<Int> {}
If you want your function to be generic then use:
如果您希望您的功能是通用的，请使用：
func myArrayFunc<T>(inputArray:Array<T>) -> Array<T> {}
If you don't want to specify type or have generic function use Anytype:
如果您不想指定类型或使用泛型函数，请使用Any类型：
func myArrayFunc(inputArray:Array<Any>) -> Array<Any> {}
回答by Valentin
Depends on what is it exactly you want to do. If you want a specialized function that takes an array of a specific type MyType, then you could write something like:
取决于你到底想做什么。如果您想要一个接受特定类型 MyType 数组的专用函数，那么您可以编写如下内容：
func myArrayFunc(inputArray: [MyType]) -> [MyType] {
// do something to inputArray, perhaps copy it?
}
If you want a generic array function, then you have to use generics. This would take an array of generic type T and return an array of generic type U:
如果你想要一个泛型数组函数，那么你必须使用泛型。这将采用泛型类型 T 的数组并返回泛型类型 U 的数组：
func myGenericArrayFunc<T, U>(inputArray: [T]) -> [U] {
}
回答by pete
thanks all (especially Kirsteins). So I've come up with this example that works well and looks logical:
谢谢大家（尤其是克尔斯泰因斯）。所以我想出了这个效果很好并且看起来合乎逻辑的例子：
func myArrayFunc(inputArray:Array<String>) -> Array<String>{
var newArray = inputArray
// do stuff with newArray
return newArray
}
回答by Renzo
There is no such thing as an Arrayin Swift, but there are arrays of a certain type, so you should give the function a generic type, like in:
Array在 Swift 中没有像 an 这样的东西，但是有某种类型的数组，所以你应该给函数一个泛型类型，比如：
func myArrayFunc<T>(inputArray:Array<T>) -> Array<T>{
// do what you want with the array
}
and then call it by instantiating T to a specific type, and passing an array of such type.
然后通过将 T 实例化为特定类型并传递此类类型的数组来调用它。
回答by oisdk
This should do it:
这应该这样做：
func myArrayFunc<T>(inputArray:Array<T>) -> Array<T> {
var newArray = inputArray
// do stuff with newArray
return newArray
}
You declare the generic type T, which is just a placeholder. Because it has no requirements Tcan be replaced by any type (when the function is called). So your function could be called like this:
您声明了泛型类型T，它只是一个占位符。因为它没有要求T可以替换为任何类型（当函数被调用时）。所以你的函数可以这样调用：
myArrayFunc([1, 2, 3])
or this:
或这个：
myArrayFunc(["a", "b", "c"])
The preferred syntax is generally [T]rather than Array<T>, though. (although both are correct)
不过，首选的语法通常是[T]而不是Array<T>。（虽然两者都是正确的）
func myArrayFunc<T>(inputArray: [T]) -> [T] {
var newArray = inputArray
// do stuff with newArray
return newArray
}
回答by halfred
Try this
尝试这个
var test = doArray([true,true,true])
test.count
func doArray(arr : [AnyObject]) -> [AnyObject] {
var _arr = arr
return _arr
}
