我正在Android中制作鼓音序器......
我正在写AudioTrack中的MODE_STREAM，以便我可以使用所有InputStreams实现同步音频播放（可通过'有效'输入流列表{{1在下面的代码中）
音频始终为：PCM（WAV），16bit Stereo 44100 Hz。
显然，我无法在UI线程上实时合成音频，因此我使用activeStreams排队所有音频缓冲。
我有缓冲回放工作，但是当合并两个（或更多）AsyncTask的缓冲区时，互联网似乎在某种关于下一步该做什么的争论中。 “将byte []转换为short []！”，“不，在运行中进行混合！”，“但是如果你不使用short，字节Endianness会被忽略！”，“它无论如何都会被忽略！” - 我甚至都不知道。
如何混合两个或更多InputStream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StackOverflow解决方案将byte []转换为short []，所以我可以将样本一起添加，但是转换总是立即崩溃Java带有一些我无法得到的神秘错误消息到处走走。所以现在我放弃了。这是我的代码实现one such StackOverflow solution ...
InputStream我目前在这一行上收到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);
}
：BufferUnderflowException。
缓冲区欠载怎么可能？我只是将byte []转换为short []。
请帮忙！
我发布了我的整个功能，希望这个更完整的代码示例和相当适应性的用法可以帮助其他人。
（P.S。字节[]到短[]转换之后是一些脆弱的硬剪辑，我甚至还没有调试，但是建议也会受到赞赏）
答案 0 :(得分：0)
你的解决方案似乎差不多好，我看到两个问题和一个潜在问题：
在这里你可以找到一个片段，我将使用两个WAV阵列（16位，单声道）的总和
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);