将short[]从SoundTouch音频库转换为回放
将short[]从SoundTouch音频库转换为回放
提问于 2014-01-05 04:39:31
我试图在安卓应用程序中使用SoundTouch C++库来实现音频速度和音高变化。我已经成功地将一个Java数组(从一个byte[] )推到JNI中，返回它，然后用一个AudioTrack回放它。
下一步是尝试将示例byte[]推送到SoundTouch管道中。我已经剖析了库中包含的SoundStretch控制台程序的源代码，并试图对其进行调整。我正在使用一个立体声，16位源，用于测试目的.
使用当前的临时设置，我忽略了RIFF头，并将其与.wav数据一起转换，因为Java对象不需要读取标头，它只播放原始的AudioTrack。在不通过byte[]发送的情况下播放原始的SoundTouch只会导致小的单击标题所在的位置。
在通过SoundTouch管道发送后，我正在播放白噪声，音频的开头应该在那里。我想我在write()函数的末尾出现了问题，在此，我要对已签名的字符进行简短的转换。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，但是我不知道如何正确地在char[]中获得这些信息，而不是写入文件。我知道我正在与演员失去信息，但我不知道如何纠正它。
如果有人有额外的动机，可以在这里找到SoundStretch源代码：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;
}程序()：
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;
}回答 1
Stack Overflow用户
回答已采纳
发布于 2014-01-05 06:01:15
我只需要获取char[]中的所有信息：
for (int i = 0; i < numElems; ++i)
{
bufferOut.push_back(temp[i] & 0xff);
bufferOut.push_back((temp[i] >> 8) & 0xff);
}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/20930127
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