ios核心音频：如何使用交织音频从AudioBuffer获取样本

Question

我使用ExtaudiofileRead函数将音频文件读入AudioBufferList。
这是音频的和ASBD：

AudioStreamBasicDescription importFormat;

importFormat.mFormatID          = kAudioFormatLinearPCM;
importFormat.mFormatFlags       = kAudioFormatFlagIsSignedInteger | kAudioFormatFlagIsPacked;
importFormat.mBytesPerPacket    = 4;
importFormat.mFramesPerPacket   = 1;
importFormat.mBytesPerFrame     = 4;
importFormat.mChannelsPerFrame  = 2;
importFormat.mBitsPerChannel    = 16;
importFormat.mSampleRate = [[AVAudioSession sharedInstance] sampleRate];

因此，我们获得并交织了2个声道的音频，每个声道的16位符号为int
AudioBufferListinit：

UInt32 *audioData = (UInt32 *) calloc (totalFramesInFile, sizeof (UInt32));

AudioBufferList *bufferList;
bufferList = (AudioBufferList *) malloc (sizeof (AudioBufferList));

// buffers amount is 1 because audio is interleaved
bufferList->mNumberBuffers = 1;

bufferList->mBuffers[0].mNumberChannels  = 2;
bufferList->mBuffers[0].mDataByteSize    = totalFramesInFile * sizeof(UInt32);
bufferList->mBuffers[0].mData            = audioData;

并读入缓冲区：

CheckError(ExtAudioFileRead (
                             audioFileObject,
                             &numberOfPacketsToRead,
                             bufferList), "error ExtAudioFileRead");

AudioFileObject是ExtaudioFileRef的和实例，它在前面的代码中启动，为了节省空间，我没有粘贴到这里。
我试图完成的是在渲染回调中修改音频样本。

OSStatus MyCallback (void *inRefCon,
                 AudioUnitRenderActionFlags *ioActionFlags,
                 const AudioTimeStamp *inTimeStamp,
                 UInt32 inBusNumber,
                 UInt32 inNumberFrames,
                 AudioBufferList *ioData){


    ViewController *view = (__bridge ViewController *) inRefCon;

    soundStruct *soundStruct = (soundStruct *) &view->mys;

    SInt64            frameTotalForSound        = soundStruct->frameCount;

    soundStruct->isPlaying = true;

    UInt32 *audioData   = soundStruct->audioData;

    UInt32 sampleNumber = soundStruct->sampleNumber;

    for( int i = 0; i < ioData->mNumberBuffers; i++){

        AudioBuffer buffer = ioData->mBuffers[i];
        UInt32 *frameBuffer = buffer.mData;

        for(UInt32 frame = 0; frame < inNumberFrames; frame++) {

            // here I fill the buffer with my audio data.
            // i need to get left and right channel samples 
            // from  audioData[sampleNumber], modify them
            // and write into frameBuffer 

            frameBuffer[frame] = audioData[sampleNumber];

            sampleNumber++;

            if(sampleNumber > frameTotalForSound) {
                soundStruct->isPlaying = false;
                AudioOutputUnitStop(soundStruct->outputUnit);
            }
        }
    }

    soundStruct->sampleNumber = sampleNumber;

    return noErr;

}

是否有可能从音频数据的UInt32阵列中获得Sint16左右声道样本？

Answer 1

让AudioData和FrameBuffer都为SINT16S:

SInt16 *audioData;
// ...
SInt16 *frameBuffer;

您的缓冲区大小计算应该是n*2*sizeof(SInt16)，您需要更改soundstruct`或添加类型转换。

然后您可以像这样访问交织的示例：

frameBuffer[0] = modify(audioData[0]);    // left sample 1
frameBuffer[1] = modify(audioData[1]);    // right sample 1
frameBuffer[2] = modify(audioData[2]);    // left sample 2
frameBuffer[3] = modify(audioData[3]);    // right sample 2
// ...
frameBuffer[2*(n-1)] = modify(audioData[2*(n-1)]);    // left sample n
frameBuffer[2*(n-1)+1] = modify(audioData[2*(n-1)+1]); // right sample n