swift:Closure声明就像块声明一样

Question

我们可以在Objective-C中声明如下所示的块.

typedef void (^CompletionBlock) (NSString* completionReason);

我试图在swift中做到这一点,它给出了错误.

func completionFunction(NSString* completionReason){ }
typealias CompletionBlock = completionFunction

错误:使用未声明的'completionFunction'

定义:

var completion: CompletionBlock = { }

这该怎么做？

更新:

根据@ jtbandes的回答,我可以创建具有多个参数的闭包

typealias CompletionBlock = ( completionName : NSString, flag : Int) -> ()

Answer 1

函数类型的语法是(in) -> out.

typealias CompletionBlock = (NSString?) -> Void
// or
typealias CompletionBlock = (result: NSData?, error: NSError?) -> Void

var completion: CompletionBlock = { reason in print(reason) }
var completion: CompletionBlock = { result, error in print(error) }

请注意,输入类型周围的括号仅在Swift 3+时需要.

Answer 2

这是关于快速关闭的精彩博客.

这里有些例子:

作为变量:

var closureName: (inputTypes) -> (outputType)

作为可选变量:

var closureName: ((inputTypes) -> (outputType))?

作为类型别名:

typealias closureType = (inputTypes) -> (outputType)