ios swift : 闭包声明就像块声明一样
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swift : Closure declaration as like block declaration
提问by Mani
We can declare block as below in Objective-C.
我们可以在 Objective-C 中如下声明块。
typedef void (^CompletionBlock) (NSString* completionReason);
I'm trying to do this in swift, it give error.
我正在尝试快速执行此操作，但会出错。
func completionFunction(NSString* completionReason){ }
typealias CompletionBlock = completionFunction
Error : Use of undeclared 'completionFunction'
错误：使用未声明的“completionFunction”
Definition :
定义 ：
var completion: CompletionBlock = { }
How to do this?
这该怎么做？
Update:
更新：
According to @jtbandes's answer, I can create closure with multiple arguments as like
根据@jtbandes 的回答，我可以像这样创建带有多个参数的闭包
typealias CompletionBlock = ( completionName : NSString, flag : Int) -> ()
回答by jtbandes
The syntax for function typesis (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) }
Note that the parentheses around the input type are only required as of Swift 3+.
请注意，输入类型周围的括号仅从 Swift 3+ 开始才需要。
回答by BLC
Hereis awesome blog about swift closure.
这是关于快速关闭的很棒的博客。
Here are some examples:
这里有些例子：
As a variable:
作为变量：
var closureName: (inputTypes) -> (outputType)
As an optional variable:
作为可选变量：
var closureName: ((inputTypes) -> (outputType))?
As a type alias:
作为类型别名：
typealias closureType = (inputTypes) -> (outputType)
