为什么Apple的闭包声明缺少参数标签
在阅读 UIKit 时，大多数时候我看到闭包（作为函数参数）缺少这样的参数标签： func fooClosure（fooClosure：（Bool）-> Swift.Void）
As I read through the UIKit, most of the time I see closures (as a function parameter) with missing argument labels like this: func fooClosure(fooClosure: (Bool) -> Swift.Void)
在某些情况下，我
看一下 UIActivityViewController 的闭包之一（作为类型别名）：
Take a look at one of UIActivityViewController's closure (as a type alias):
public typealias UIActivityViewControllerCompletionWithItemsHandler = (UIActivityType?, Bool, [Any]?, Error?) -> Swift.Void
3rd 类型代表什么？我必须查看代码的 Objective-C 版本，才能知道这些是 returnedItems 。
What does 3rd type stand for? I have to look at the Objective-C version of the code to know these are returnedItems.
为什么不这样实现：（请注意 activityItems 标签）
Why it is not implemented that way: (note the activityItems label)
public typealias UIActivityViewControllerCompletionWithItemsHandler = (UIActivityType?, Bool, _ activityItems: [Any]?, Error?) -> Swift.Void
编辑：
-是这只是一个临时状态（因为Swift尚未完全集成到这些框架中）？
没有参数标签的代码可读性较低。
问题是这是一个函数签名机器生成的typealias，机器没有按照您的意愿执行操作。考虑下面的Objective-C声明：
The problem is that this is a function signature typealias generated by machine, and the machine isn't doing what you would like it to do. Consider the following Objective-C declaration:
typedef void (^MyCoolType)(NSString* __nullable name);
Swift生成的界面是：
The Swift generated interface for that is:
public typealias MyCoolType = (String?) -> Swift.Void
如您所见，名称被剥夺；
在Swift中，没有什么可以阻止 you 定义带有内部标签的函数签名类型别名的。这是合法的：
Nothing stops you from defining a function signature typealias with internal labels, in Swift. This is legal:
public typealias MyCoolType = (_ name:String?) -> Void
但这并不是从Objective-C生成生成接口的方式。
But that does not happen to be how the generated interface is generated from Objective-C.