调用可以抛出,但不能从全局变量初始化程序中抛出错误

Question

我正在使用Xcode 7 beta,在迁移到Swift 2后,我遇到了一些代码问题:

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

我得到一个错误说"调用可以抛出,但错误不能从全局变量初始化程序中抛出".我的应用程序依赖于它recorder是一个全局变量.有没有办法让它保持全球化,但解决这些问题？我不需要高级错误处理,我只是想让它工作.

Answer 1

如果您知道函数调用不会抛出异常,则可以调用throw函数try!来禁用错误传播.请注意,如果实际抛出错误,这将抛出运行时异常.

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

来源:Apple错误处理文档(禁用错误传播)

Answer 2

您可以使用3种方法来解决此问题.

• 使用try创建可选的AVAudioRecorder？
• 如果你知道它会让你回归AVRecorder,你可以谨慎使用试试!
• 或者使用try/catch处理错误

用试试？

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

用试试看!

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

试着抓

// 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)")
}