InputStream 混音(MODE_STREAM)

Question

我正在 Android...

制作鼓音序器

我正在写入 MODE_STREAM 中的 AudioTrack，这样我就可以实现与所有 InputStreams 的同步音频播放（可通过 'active' InputStreams 列表获得, activeStreams 在下面的代码中)

音频始终为：PCM (WAV)，16 位立体声 44100 Hz。

显然，我无法在 UI 线程上实时合成音频，所以我使用 AsyncTask 来排队所有音频缓冲。

我的缓冲回放可以正常工作，但是当谈到合并两个（或更多）InputStream 的缓冲区时，互联网似乎在讨论下一步该怎么做。 "Convert the byte[] to short[]!"、"No, do the bit mixing on-the-fly!"、"But if you don't use shorts the byte Endianness is ignored!"、"It gets ignored anyway!" - 我什至不知道了。

如何混合两个或多个 InputStreams 的缓冲区？我不明白为什么我当前的实现失败

我试过 4 种不同的 Whosebug 解决方案来将 byte[] 转换为 short[]，这样我就可以将样本加在一起，但是转换总是立即崩溃 Java 并出现一些神秘的错误消息我无法理解。所以现在我放弃了。这是我的代码实现 one such Whosebug solution...

protected Long doInBackground(Object ... Object) {

  int bytesWritten = 0;
  InputStream inputStream;
  int si = 0, i = 0;

  //The combined buffers. The 'composition'
  short[] cBuffer = new short[Synth.AUDIO_BUFFER_SIZE];

  //The 'current buffer', the segment of inputStream audio.
  byte[] bBuffer = new byte[Synth.AUDIO_BUFFER_SIZE];

  //The 'current buffer', converted to short?
  short[] sBuffer = new short[Synth.AUDIO_BUFFER_SIZE];

  int curStreamNum;
  int numStreams = activeStreams.size();
  short mix;

  //Start with an empty 'composition'
  cBuffer = new short[Synth.AUDIO_BUFFER_SIZE];

  boolean bufferEmpty = false;
  try {
    while(true) { // keep going forever, until stopped or paused.
      for(curStreamNum = 0;curStreamNum < numStreams;curStreamNum++){
        inputStream = activeStreams.get(curStreamNum);
        i = inputStream.read(bBuffer);
        bufferEmpty = i<=-1;
        if(bufferEmpty){
          //Input stream buffer was empty. It's out of audio. Close and remove the stream.
          inputStream.close();
          activeStreams.remove(curStreamNum);
          curStreamNum--; numStreams--; continue; // hard continue.
        }else{
          //Take the now-read buffer, and convert to shorts.
          ByteBuffer.wrap(bBuffer).order(ByteOrder.LITTLE_ENDIAN).asShortBuffer().get(sBuffer);

          //Take the short buffer, merge into composition buffer.
          //TODO: Optimize by making the 'first layer' of the composition the first buffer, on its own.
          for(si=0;si<Synth.AUDIO_BUFFER_SIZE;si++){
            mix = (short) (sBuffer[si] + cBuffer[si]);
            //This part is probably completely wrong too. I'm not up to here yet to evaluate whats needed...
            if(mix >= 32767){
              mix = 32767;
            }else if (mix <= -32768){
              mix = -32768;
            }
            cBuffer[si] = mix;
          }
        }
      }
      track.write(sBuffer, 0, i);

      //It's always full; full buffer of silence, or of composited audio.
      totalBytesWritten += Synth.AUDIO_BUFFER_SIZE;

      //.. queueNewInputStreams ..
      publishProgress(totalBytesWritten);
      if (isCancelled()) break;
    }
  } catch (IOException e) {e.printStackTrace();}
  return Long.valueOf(totalBytesWritten);
}

我目前在这条线上收到 BufferUnderflowException：ByteBuffer.wrap(bBuffer).order(ByteOrder.LITTLE_ENDIAN).asShortBuffer().get(sBuffer);。

怎么可能出现缓冲区欠载？我只是将 byte[] 转换为 short[].

请帮忙！

我已经发布了我的整个函数，希望这个更完整的代码示例和相当适应性强的用法可以帮助其他人。

(P.S。byte[] 到 short[] 的转换之后是一些脆弱的硬剪辑，我什至还没有调试，但也将不胜感激)

Answer 1

您的解决方案似乎不错，我看到两个问题和一个潜在问题：

短数组的长度：必须是字节数组的一半，否则会下溢
短裤的总和必须是短裤的平均值而不仅仅是总和，否则你只会得到噪音
（潜在问题）您通过 InputStream 读取的数组长度不能完全免费，因为您必须为每个 InputStream 求和 2 个字节（那么它必须是一个偶数数组）并且您应该注意单声道与单声道。立体声音频文件（如果立体声，左声道有 2 个字节，右声道有 2 个字节交错）

在这里你可以找到一个片段，我会用它来求和两个 WAV 数组（16 位，单声道）

    Random random = new Random();

    int bufferLength = 20;

    byte[] is1 = new byte[bufferLength];
    byte[] is2 = new byte[bufferLength];
    byte[] average = new byte[bufferLength];

    random.nextBytes(is1);
    random.nextBytes(is2);

    short[] shorts1 = new short[bufferLength/2];
    ByteBuffer.wrap(is1).order(ByteOrder.LITTLE_ENDIAN).asShortBuffer().get(shorts1);

    short[] shorts2 = new short[bufferLength/2];
    ByteBuffer.wrap(is2).order(ByteOrder.LITTLE_ENDIAN).asShortBuffer().get(shorts2);

    short[] result = new short[bufferLength/2];

    for (int i=0; i<result.length; i++) {
        result[i] = (short) ((shorts1[i] + shorts2[i])/2);
    }

    ByteBuffer.wrap(average).order(ByteOrder.LITTLE_ENDIAN).asShortBuffer().put(result);

对于 32 位立体声，解决方案可能是

    Random random = new Random();

    int bufferLength = 8 * 50;

    byte[] is1 = new byte[bufferLength];
    byte[] is2 = new byte[bufferLength];
    byte[] average = new byte[bufferLength];

    random.nextBytes(is1);
    random.nextBytes(is2);

    System.out.println(bytesToHex(is1));
    System.out.println(bytesToHex(is2));

    int[] ints1 = new int[bufferLength/4];
    ByteBuffer.wrap(is1).order(ByteOrder.LITTLE_ENDIAN).asIntBuffer().get(ints1);

    int[] ints2 = new int[bufferLength/4];
    ByteBuffer.wrap(is2).order(ByteOrder.LITTLE_ENDIAN).asIntBuffer().get(ints2);

    int[] result = new int[bufferLength/4];

    for (int i=0; i<result.length; i++) {
        result[i] = ((ints1[i] + ints2[i])/2);
    }

    ByteBuffer.wrap(average).order(ByteOrder.LITTLE_ENDIAN).asIntBuffer().put(result);

InputStream 混音(MODE_STREAM)

InputStream audio mixing (MODE_STREAM)

java

android

audio

buffer

inputstream