在快速closures中返回void和（）之间的区别

这些closures有什么区别？

let closureA: () -> () 

和

 let closureB: () -> Void 

Answer 1

如果你看看标题，你会看到Void是（）的typealias，

 /// The empty tuple type. /// /// This is the default return type of functions for which no explicit /// return type is specified. typealias Void = () 

你可以validation使用这样的操场，

 let a = Void() println(a) // prints () 

更新

如果你想在头文件中看到Void的声明，在上面的代码中，在swift的操作系统或xcode编辑器中input如下代码，

  let a: Void = () 

然后，cmd +在上面的expression式中点击“Void”关键字，你会被带到头文件，在那里你可以看到Void的声明。

这个文件已经更新了更多的信息，就是这样，

 /// The return type of functions that don't explicitly specify a return type; /// an empty tuple (ie, `()`). /// /// When declaring a function or method, you don't need to specify a return /// type if no value will be returned. However, the type of a function, /// method, or closure always includes a return type, which is `Void` if /// otherwise unspecified. /// /// Use `Void` or an empty tuple as the return type when declaring a /// closure, function, or method that doesn't return a value. /// /// // No return type declared: /// func logMessage(_ s: String) { /// print("Message: \(s)") /// } /// /// let logger: (String) -> Void = logMessage /// logger("This is a void function") /// // Prints "Message: This is a void function" public typealias Void = () 

Answer 2

相同。 这只是一个typealias，所以它的工作原理是一样的。

 typealias Void = () 

听起来像埃里卡·萨登和苹果正试图坚持虚无： http : //ericasadun.com/2015/05/11/swift-vs-void/

Answer 3

根本没有区别

Void是（）的别名：

 typealias Void = ()