从 SoundTouch 音频库转换一个 short[] 以进行播放

Question

我正在尝试使用 SoundTouch C++ 库来实现 Android 应用程序中的音频速度和音高变化。我已经成功地通过 JNI 推送了一个 Java byte[] 数组（来自 .wav），将其返回，并使用 AudioTrack 播放它。

下一步是尝试通过 SoundTouch 管道推送样本字节 []。我已经剖析了库中包含的 SoundStretch 控制台程序的源代码，并尝试对其进行调整。我正在使用立体声 16 位源进行测试。

使用我当前的临时设置，我将忽略 RIFF 标头并将其与 .wav 数据一起转换，因为 Java AudioTrack 对象不需要读取标头，它只播放原始 PCM。在不通过 SoundTouch 发送的情况下播放原始字节 [] 只会导致在标题所在的位置单击一下。

通过 SoundTouch 管道发送后，我正在播放应该是音频开头的白噪声。我假设我在write() 函数的末尾遇到了问题，我在其中将short's 转换为签名字符。在这里，控制台应用程序正在写入文件，而不是推送到向量：

int res = (int)fwrite(temp, 1, numBytes, fptr);

我已经阅读了fwrite 的文档，但我对位旋转或音频处理知之甚少，无法知道如何在 char[] 中正确获取此信息而不是写入文件。我知道我正在丢失有关演员的信息，但我不确定如何更正它。

如果有人有额外的动力，可以在此处找到 SoundStretch 源：http://www.surina.net/soundtouch/sourcecode.html

extern "C" DLL_PUBLIC jbyteArray 
Java_net_surina_soundtouch_SoundTouch_getMutatedBytes
(JNIEnv *env, jobject thiz, jbyteArray input, jint length)
{
    const int BUFF_SIZE = 2048000;

    SoundTouch soundTouch;

    jboolean isCopy;
    jbyte* ar = env->GetByteArrayElements(input, &isCopy);

    signed char* cBufferIn = (signed char*)ar;

    SAMPLETYPE* fBufferIn = new SAMPLETYPE[length];
    vector<signed char> fBufferOut;

    //converts the chars to floats per the SoundTouch console app.
    convertInput16(cBufferIn, fBufferIn, length); 

    //channels, sampling rate, speed, pitch change
    setup(&soundTouch, 2, 44100, 1.0, 0);

    //transform floats from fBufferIn to fBufferout
    process(&soundTouch, fBufferIn, fBufferOut, BUFF_SIZE); 

    signed char* res = &fBufferOut[0];

    jbyteArray result = env->NewByteArray(length);
    env->SetByteArrayRegion(result, 0, fBufferOut.size(), res);

    LOGV("fBufferOut Size: %d", fBufferOut.size());

    delete[] fBufferIn;

    return result;
}

进程（）：

static void process(SoundTouch* soundTouch, SAMPLETYPE* fBufferIn, vector<signed char>& fBufferOut, int BUFF_SIZE)
{
    int nSamples = BUFF_SIZE / 2; //2 bytes per sample, using 16 bit sample for testing
    int buffSizeSamples = BUFF_SIZE / 2; //2 channel stereo

    soundTouch->putSamples(fBufferIn, nSamples);

    do
    {
        nSamples = soundTouch->receiveSamples(fBufferIn, buffSizeSamples);
        write(fBufferIn, fBufferOut, nSamples / 2); //2 channels
    } while (nSamples != 0);

    soundTouch->flush();

    do
    {
        nSamples = soundTouch->receiveSamples(fBufferIn, buffSizeSamples);
        write(fBufferIn, fBufferOut, nSamples / 2);
        LOGV("NUMBER OF SAMPLES: %d", nSamples);
    } while (nSamples != 0);
}

写（）：

static void write(const float *bufferIn, vector<signed char>& bufferOut, int numElems)
{
    int numBytes;
    int bytesPerSample;

    if (numElems == 0) return;

    bytesPerSample = 16 / 8; //16 bit test sample / bits in a byte
    numBytes = numElems * bytesPerSample;
    short *temp = (short*)getConvBuffer(numBytes);

    switch (bytesPerSample)
    {

    case 2: //16 bit encoding per the SoundStretch console app
    {
              short *temp2 = (short *)temp;
              for (int i = 0; i < numElems; i++)
              {
                  short value = (short)saturate(bufferIn[i] * 32768.0f, -32768.0f, 32767.0f); //magic to me
                  temp2[i] = value; //works for little endian only.
              }
              break;
    }

    default:
        assert(false);
    }

    for (int i = 0; i < numElems; ++i)
    {
        bufferOut.push_back((signed char)temp[i]); //I think my problem is here.
    }

    delete[] temp;

    //bytesWritten += numBytes;
}

Answer 1

我只需要获取 char[] 中的所有位：

for (int i = 0; i < numElems; ++i)
{
    bufferOut.push_back(temp[i] & 0xff);
    bufferOut.push_back((temp[i] >> 8) & 0xff);
}