ios - 无法在 Swift xcode 6.3.1 中使用类型为 'functionName' 的参数列表调用 '([(nameOfClass)])'

Question

我现在正在学习 swift 并遇到以下问题请帮忙..

我有3节课- TableView Controller 广播模型广播请求我收到以下错误(错误行已用注释标记) 无法使用“([(BroadcastModel)])”类型的参数列表调用“requestFinished”

import UIKit

public class TableViewController: UITableViewController {

var broadcasts = [BroadcastModel]()
//MARK: ViewControllerLifecycle
override public func viewDidLoad() {

    super.viewDidLoad()
    //maybe will use the 2d array for sections of broadcasts..
    BroadcastRequest().requestNewBroadcasts()

}

public func requestFinished(requestedBroadcasts: [BroadcastModel]) {
    self.broadcasts = requestedBroadcasts  \* HERE IS THE ERROR *\
    self.tableView.reloadData()
}

public class BroadcastRequest {

func requestNewBroadcasts() {
    var broadcasts = [BroadcastModel]()
    .....
    .....
    broadcasts.append(broadcast)
    TableViewController.requestFinished(broadcasts)
}
}

public class BroadcastModel: NSObject, Printable {
let id: String
let broadcastURL: String
...
...
override public var description: String {
    return "ID: \(id), URL: \(broadcastURL) ....."
}

init(...) {
... 
}
}

Answer 1

因为您正在使用:

 TableViewController.requestFinished(broadcasts)

您应该将函数定义为类函数:

class func requestFinished(requestedBroadcasts: [BroadcastModel]) {
    self.broadcasts = requestedBroadcasts  \* HERE IS THE ERROR *\
    self.tableView.reloadData()
}

建议大家搞清楚函数和类函数的区别，类本身和类实例的区别

如果你想对一个类实例做一些事情，你必须有它的引用，而不是只有类名。