ios - 快速从另一个 ViewController 调用函数

Question

我已经查看了 Stackoverflow，但无法得到答案。我想创建停止在另一个 ViewController 中播放声音的功能。但是当我点击停止按钮时，它崩溃了并显示“EXC_BAD_INSTRUCTION (code=EXC_I386_INVOP, subcode=0x0)”。这是我的代码。

第一个 View Controller

import UIKit
import AVFoundation

class FirstVC: UIViewController {

   var metronome: AVAudioPlayer!
   override func viewDidLoad() {
       super.viewDidLoad()
   do {
        let resourcePath1 = Bundle.main.path(forResource: "music", ofType: "mp3")
        let url = NSURL(fileURLWithPath: resourcePath1!)
        try metronome = AVAudioPlayer(contentsOf: url as URL)

        metronome.prepareToPlay()
        metronome.play()
    } catch let err as NSError {
        print(err.debugDescription)
    }
}

另一个Viewcontroller是

import UIKit
class SecondVC: UIViewController {
   var metronomePlay = FirstVC()

@IBAction func stopBtnPressed(_ sender: Any) {
   metronomePlay.metronome.stop() //"EXC_BAD_INSTRUCTION (code=EXC_I386_INVOP, subcode=0x0)"
   }
}

Answer 1

从今天的 swift 4.1 开始，这段代码对我有用:

将其放入发送 Controller 中:

NotificationCenter.default.post(name: Notification.Name(rawValue: "disconnectPaxiSockets"), object: nil)

将其放入接收 Controller viewDidLoad() 或 viewWillAppear() 中:

NotificationCenter.default.addObserver(self, selector: #selector(disconnectPaxiSocket(_:)), name: Notification.Name(rawValue: "disconnectPaxiSockets"), object: nil)

然后是接收 Controller 类中的以下函数:

@objc func disconnectPaxiSocket(_ notification: Notification) {
    ridesTimer.invalidate()
    shared.disconnectSockets(socket: self.socket)
}