快速回调函数

Question

我正在尝试使用 The Amazing Audio Engine在 OS X 上使用 Swift 录制一些音频。为此，我需要实现一个回调函数来接收音频并对其进行处理。该文档有一些 examples关于如何使用 Objective-C 做到这一点:

@interface MyAudioReceiver : NSObject <AEAudioReceiver>
@end
@implementation MyAudioReceiver
static void receiverCallback(__unsafe_unretained MyAudioReceiver *THIS,
 __unsafe_unretained AEAudioController *audioController,
 void *source,
 const AudioTimeStamp *time,
 UInt32 frames,
 AudioBufferList *audio) {

 // Do something with 'audio'
}
-(AEAudioReceiverCallback)receiverCallback {
 return receiverCallback;
}
@end
...
id<AEAudioReceiver> receiver = [[MyAudioReceiver alloc] init];

或

id<AEAudioReceiver> receiver = [AEBlockAudioReceiver audioReceiverWithBlock:
 ^(void *source,
 const AudioTimeStamp *time,
 UInt32 frames,
 AudioBufferList *audio) {
 // Do something with 'audio'
}];

据我所知:

var audioController: AEAudioController? = nil
audioController = AEAudioController(audioDescription: AEAudioStreamBasicDescriptionInterleaved16BitStereo, inputEnabled: true)
do {
    try audioController?.start()
} catch {
    NSLog("An error happened while starting AEAudioController.")
}

let receiver = MyAudioReceiver();
audioController?.addInputReceiver(receiver)

class MyAudioReceiver : NSObject, AEAudioReceiver {
    var receiverCallback: AEAudioReceiverCallback! {
        // what do I do here?
    }
}

现在我在 receiverCallback 属性中遇到错误。我是在正确的轨道上还是我的方法完全错误？

我不知道如何在 Swift 中做完全相同的事情。我该怎么做？

Answer 1

在 Swift 中，函数和 block 几乎是平等对待的，统一的概念是 closures .我建议您阅读(连同文档的其余部分)以了解闭包语法和语义。