Call can throw, but errors can not be thrown out of a global variable initializer

Question

I'm using Xcode 7 beta and after migrating to Swift 2 I experienced some issues with this line of code:

let recorder = AVAudioRecorder(URL: soundFileURL, settings: recordSettings as! [String : AnyObject])

I get an error saying "Call can throw, but errors can not be thrown out of a global variable initializer". My app relies on the recorder to be a global variable. Is there a way to keep it global but fix these issues? I have no need for advanced error handling, I just want it to work.

Answer 1

If you know that your function call will not throw an exception, you can call the throwing function with try! to disable error propagation. Note that this will throw a runtime exception if an error is actually thrown.

let recorder = try! AVAudioRecorder(URL: soundFileURL, settings: recordSettings as! [String : AnyObject])

Source: Apple Error Handling documentation (Disabling Error Propagation)

Answer 2

There are 3 ways that you can use to solve this problem.

• Creating optional AVAudioRecorder using try?
• If you know that it will return you AVRecorder, you can implicity use try!
• Or then handle the error using try / catch

Using try?

// notice that it returns AVAudioRecorder?
if let recorder = try? AVAudioRecorder(URL: soundFileURL, settings: recordSettings) { 
    // your code here to use the recorder
}

Using try!

// this is implicitly unwrapped and can crash if there is problem with soundFileURL or recordSettings
let recorder = try! AVAudioRecorder(URL: soundFileURL, settings: recordSettings)

try / catch

// The best way to do is to handle the error gracefully using try / catch
do {
    let recorder = try AVAudioRecorder(URL: soundFileURL, settings: recordSettings)
} catch {
    print("Error occurred \(error)")
}