android - 如何在android中将8位和16位pcm文件合并到单独的音频 channel 中

Question

我知道有很多类似的问题，我阅读了很多，但我无法让它发挥作用。正如我在标题中所说，我有一个 8 位和一个 16 位 pcm 单声道文件，我想将它们合并到一个立体声波文件中，但在不同的 channel 中，一个在左 channel ，一个在右 channel 。

到目前为止，我已经完成了以下操作，但无论我做什么，都会出现两个问题:
1-我不能将它们写在输出文件的单独 channel 中。
2- 其中一个文件会写得像噪音(“RECORDER_BPP”是 8 还是 16)

我的代码:

new Thread(new Runnable() {
        @Override
        public void run() {
            int channels = 2;
            RECORDER_BPP = 16;
            int RECORDER_SAMPLERATE = 44100;
            int minBufferSize = AudioTrack.getMinBufferSize(maxCaptureRate, AudioFormat.CHANNEL_IN_MONO, AudioFormat.ENCODING_PCM_16BIT);

            byte[] data = new byte[minBufferSize];
            long longSampleRate = RECORDER_SAMPLERATE;
            long byteRate = RECORDER_BPP * RECORDER_SAMPLERATE * channels/8;

            try {
                FileInputStream in1 = new FileInputStream(rawAudio1);
                FileInputStream in2 = new FileInputStream(rawAudio2);
                FileOutputStream out = new FileOutputStream(wavAudio);
                long totalAudioLen = in1.getChannel().size();
                long totalDataLen = totalAudioLen + 36;

                WriteWaveFileHeader(out, totalAudioLen, totalDataLen,
                        longSampleRate, channels, byteRate);

                while(in1.read(data) != -1){
                    out.write(data);

                    //2nd channel ?! // unsuccessful
                    in2.read(data);
                    out.write(data);
                }

                in1.close();
                in2.close();
                out.close();
            } catch (Exception e) {
                Log.d("EEE-2ch", e.getMessage());
            }
        }
}).start();

这是标题部分:

private void WriteWaveFileHeader(
        FileOutputStream out, long totalAudioLen,
        long totalDataLen, long longSampleRate, int channels,
        long byteRate) throws IOException {

    byte[] header = new byte[44];

    header[0] = 'R';  // RIFF/WAVE header
    header[1] = 'I';
    header[2] = 'F';
    header[3] = 'F';
    header[4] = (byte) (totalDataLen & 0xff);
    header[5] = (byte) ((totalDataLen >> 8) & 0xff);
    header[6] = (byte) ((totalDataLen >> 16) & 0xff);
    header[7] = (byte) ((totalDataLen >> 24) & 0xff);
    header[8] = 'W';
    header[9] = 'A';
    header[10] = 'V';
    header[11] = 'E';
    header[12] = 'f';  // 'fmt ' chunk
    header[13] = 'm';
    header[14] = 't';
    header[15] = ' ';
    header[16] = 16;  // 4 bytes: size of 'fmt ' chunk
    header[17] = 0;
    header[18] = 0;
    header[19] = 0;
    header[20] = 1;  // format = 1
    header[21] = 0;
    header[22] = (byte) channels;
    header[23] = 0;
    header[24] = (byte) (longSampleRate & 0xff);
    header[25] = (byte) ((longSampleRate >> 8) & 0xff);
    header[26] = (byte) ((longSampleRate >> 16) & 0xff);
    header[27] = (byte) ((longSampleRate >> 24) & 0xff);
    header[28] = (byte) (byteRate & 0xff);
    header[29] = (byte) ((byteRate >> 8) & 0xff);
    header[30] = (byte) ((byteRate >> 16) & 0xff);
    header[31] = (byte) ((byteRate >> 24) & 0xff);
    header[32] = (byte) (channels * RECORDER_BPP / 8);  // block align
    header[33] = 0;
    header[34] = (byte) RECORDER_BPP;  // bits per sample
    header[35] = 0;
    header[36] = 'd';
    header[37] = 'a';
    header[38] = 't';
    header[39] = 'a';
    header[40] = (byte) (totalAudioLen & 0xff);
    header[41] = (byte) ((totalAudioLen >> 8) & 0xff);
    header[42] = (byte) ((totalAudioLen >> 16) & 0xff);
    header[43] = (byte) ((totalAudioLen >> 24) & 0xff);

    out.write(header, 0, 44);
}

感谢您的帮助，谢谢

Answer 1

输出数据需要交错，这意味着您需要从 in1 读取单个样本，将其写入输出，然后从 in2 读取单个样本并将其写入输出。由于 in2 是 8 位音频，您需要读取 8 位样本，然后写出 16 位。

byte[] data = new byte[2];
while(in1.read(data) != -1) { // read 2 bytes
    out.write(data);          // write 2 bytes (L)
    in2.read(data, 0, 1);     // read 1 byte
    data[1] = 0;              // set the other byte to zero
    // if this is the wrong byte order then do this
    // data[1] = data[0];
    // data[0] = 0;
    out.write(data);          // write 2 bytes (R)
}

此代码假定 in1 是具有 16 位音频的文件。如果那是错误的，那么您将需要对其进行一些修改。