CODEWITHSUNDEEP
iosxcodeswiftavaudioplayer
Alexander Khitev
Alexander Khitev

Reputation: 6841

When I stop AVAudioPlayer I get an error EXC_BAD_ACCESS

I have a UIViewController which is playing music and UITableViewController which lists songs. I made that I leave PlayViewController, AVAudioPlayer will still be playing a song. I want that I return on UITableViewController and I choose a different song the AVAudioPlayer will be stopped. I tried many different methods but I get the same error EXC_BAD_ACCESS. I also made AVAudioPlayer like weak var and I also made PlayViewController is a public class but I got the same error. How can I to fix this problem?

        override func prepareForSegue(segue: UIStoryboardSegue, sender: AnyObject?) {
        if segue.identifier == "playMusic" {
            var playMVC = (segue.destinationViewController as! UINavigationController).topViewController as! PlayMusicVC
            var currentIndPass = tableView.indexPathForSelectedRow()!.row
//            println("array item \(arrayItem)")
            playMVC.currentIndex = currentIndPass
            playMVC.arrayOfSongs = arrayItem
            searchController.active = false
//            println("array item \(arrayItem)")
            //test
            var secondPlayView = PlayFromSearchVC()
            if secondPlayView.audioPlayer.playing == true {
                secondPlayView.audioPlayer.stop()
                println("It is works")
            } else {
                println("It is not works")
            }
            //test
        } else if segue.identifier == "summer" {
//            println("array super filter and count \(arrayFilterOfNames) co \(arrayFilterOfNames.count)")
            var playVC = (segue.destinationViewController as! UINavigationController).topViewController as! PlayFromSearchVC
            var currentInt = tableView.indexPathForSelectedRow()!.row
            playVC.currentSongString = arrayFilterOfNames[currentInt]
//            playVC.currentIndex = currentInt
//            playVC.arrayOfSongs = arrayFilterOfNames // arrayFilterNames
            searchController.active = false
            //
            var secondPlayView = PlayFromSearchVC()
            if secondPlayView.audioPlayer.playing == true {
                secondPlayView.audioPlayer.stop()
                println("It is works")
            } else {
                println("It is not works")
            }
            //
        }
    }

AVAudioPlayer is 

    var audioPlayer = AVAudioPlayer()

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Different variant is wrong too enter image description here

Upvotes: 0

Views: 973

Answers (3)

WINSergey
WINSergey

Reputation: 2005

When you try to set currentTime to AVAudioPlayer item but before you set playing content -> you'll definitely get such error

Upvotes: 0

Duncan C
Duncan C

Reputation: 131398

This code makes no sense:

var secondPlayView = PlayFromSearchVC()
if secondPlayView.audioPlayer.playing == true {

The first line creates a new instance of PlayFromSearchVC. Since it did not exist before, the second line does not make sense. It's like asking a baby that was just born about it's favorite toy.

You don't show the structure of your PlayViewController class. How is audioPlayer defined?

Don't use a view controller that can be closed to manage sound play. Create a singleton object that manages sound play. Add a method to the sound manager that stops any sound that is currently playing, and methods to start a new sound playing. Call the sound manager from everywhere that you need to play sounds.

Upvotes: 1

Kalaivani
Kalaivani

Reputation: 424

Since the AvAudioPlayer (secondPlayView) is a weak property.It would have gone while PlayViewController is removed from memory.Make sure your AVAudioPlayer instance is initialized properly or referring the same instance when you come back to PlayViewController.

Upvotes: 1

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