swift - 调用以 TypeAlias 作为参数的函数？

Question

所以我写了一个与闭包有关的小练习程序。我试图更好地理解异步概念是如何工作的。当我尝试调用 request() 时，出现如下所示的转换错误:

import UIKit

let correctPasscode = "3EyX"

typealias CompletionHandler = (result: AnyObject?, error: String?) -> Void

func request(passcode: String, completionHandler: CompletionHandler) {

sendBackRequest(passcode) {(result, error) -> Void in
    if error != nil {
        print(error)
    }
    else {
        print(result)

    }}

}

func sendBackRequest(passCode: String, completionHandler: CompletionHandler) {
if passCode == correctPasscode {
    completionHandler(result: "Correct. Please proceed", error: nil)

} else {
    completionHandler(result: nil, error: "There was an error signing in")

}
}


request(correctPasscode, completionHandler: CompletionHandler) // Error happens here

Answer 1

类型别名告诉您需要传递什么实际 类型。在这种情况下，类型是 闭包 类型

(result: AnyObject?, error: String?) -> Void

你像这样传递它:

request(correctPasscode, completionHandler:{
    (result: AnyObject?, error: String?) in 
    print("Inside the handler...")
    // Do some useful things here
})

甚至更短-

request(correctPasscode) {
    (result: AnyObject?, error: String?) in 
    print("Inside the handler...")
    // Do some useful things here
}

或者更短——(类型通过 func 声明已知)——

request(correctPasscode) { result, error in
    print("Inside the handler...")
    // Do some useful things here
}