'return' 之后的代码永远不会被执行

Question

我正处于编码的第一周。 到目前为止，我在 swift 中制作的 youtube 教程应用程序与上述错误消息完全不同。这是代码。我现在完全按照他们的规定重写了两次。 我正在尝试设置声音以与我们正在制作的纸牌游戏一起玩，当纸牌被洗牌、转身、匹配或不正确匹配时播放的声音。 错误消息显示为“let soundURL”代码行，“'return' 之后的代码将永远不会被执行”。 请帮忙？

class SoundManager {

   static var audioPlayer:AVAudioPlayer?

    enum SoundEffect {

        case flip
        case shuffle
        case match
        case nomatch

    }

   static func playSound(_ effect:SoundEffect) {

        var soundFilename = ""

        // Determine which sound effect we want to play
        //and set the appropriate file name

        switch effect {

        case .flip:
            soundFilename = "cardflip"

        case .shuffle:
            soundFilename = "cardflip"

        case .match:
            soundFilename = "dingcorrect"

        case.nomatch:
            soundFilename = "dingwrong"

        }
        // Get the path to the sound file inside the bundle
        let bundlePath = Bundle.main.path(forResource: soundFilename, ofType: "wav")

        guard bundlePath != nil else {
            print("Couldn't find sound file \(soundFilename) in the bundle")
            return



            // Create a URL object from this string path
            let soundURL = URL(fileURLWithPath: bundlePath!)


            do {
                // Create audio player object
                audioPlayer = try AVAudioPlayer(contentsOf: soundURL)

                // Play the sound
                audioPlayer?.play()
            }
            catch {
                // Could'nt create audio player object, log the error
                print("Could'nt create the audio player object for sound file \(soundFilename)")
            }


        }

    }
}

Answer 1

我相信您希望您的代码看起来像这样：

// Get the path to the sound file inside the bundle
let bundlePath = Bundle.main.path(forResource: soundFilename, ofType: "wav")

guard bundlePath != nil else {
    print("Couldn't find sound file \(soundFilename) in the bundle")
    return
}

// Create a URL object from this string path
let soundURL = URL(fileURLWithPath: bundlePath!)


do {
    // Create audio player object
    audioPlayer = try AVAudioPlayer(contentsOf: soundURL)

    // Play the sound
    audioPlayer?.play()
}
catch {
    // Could'nt create audio player object, log the error
    print("Could'nt create the audio player object for sound file \(soundFilename)")
}

如果有一个guard 语句，其括号内有return，您不想在return 下方和相同的括号内放置任何代码。 return 下面的代码都不会执行。在我的示例中，以下代码位于括号之外，这意味着只要有bundlePath 的值，它就会执行。

Answer 2

Code after 'return' will never be executed



var soundFilename = ""

// Determine which sound effect we want to play
//and set the appropriate file name

switch effect {

case .flip:
    soundFilename = "cardflip"

case .shuffle:
    soundFilename = "cardflip"

case .match:
    soundFilename = "dingcorrect"

case.nomatch:
    soundFilename = "dingwrong"

}
// Get the path to the sound file inside the bundle
let bundlePath = Bundle.main.path(forResource: soundFilename, ofType: "wav")

guard bundlePath != nil else {
    print("Couldn't find sound file \(soundFilename) in the bundle")
    return

   }

    // Create a URL object from this string path
    let soundURL = URL(fileURLWithPath: bundlePath!)


    do {
        // Create audio player object
        audioPlayer = try AVAudioPlayer(contentsOf: soundURL)

        // Play the sound
        audioPlayer?.play()
    }
    catch {
        // Could'nt create audio player object, log the error
        print("Could'nt create the audio player object for sound file \(soundFilename)")
    }
// }

}

我相信您误解了保护语法，因为它会在条件正确时执行代码，而在条件为假时会落入 else 部分。因此，它将在没有其他代码的情况下在 else 中使用 return，这样该方法将优雅地终止而不会造成任何崩溃。

return 关键字用于在方法执行时返回某些内容或退出方法。

所以你需要用“}”关闭else部分并删除catch块下面的一个。希望这会有所帮助。