AudioRecord：如何使用常见的缓冲区来使用它用于处理和存储？ [英] AudioRecord : How can I use a common buffer to use it for processing and storing?
问题描述
我有写入数据库中的 AudioRecord 线程。现在，我想在一些间隔缓冲区使用一些音频数据，并利用处理它的 FFT 。我想送音频缓冲区的 FFT 作为参数。
I have an AudioRecord thread that writes to database. Now I want to use some audio data in buffer at some intervals and process it using FFT. I want to send audio buffer to FFT as parameter.
当我试图用一个共同的缓冲区，它给我的的libc 错误。我如何使用通用缓冲将它传递给 FFT ，并把它写入存储？
When I am trying to use a common buffer then its giving me libc error. How can I use a common buffer to pass it to FFT and also write it to a storage?
当我尝试使用不同的读取调用有数据丢失的情况，因此不能使用了。
When I tried using different read calls there was situation of data loss and hence cant use that.
以下是我的code
public void start() {
startRecording();
_isRecording = true;
_recordingThread = new Thread(new Runnable() {
public void run() {
writeAudioDataToFile();
}
}, "AudioRecorder Thread");
_recordingThread.start();
}
private void writeAudioDataToFile() {
while (_isRecording) {
// gets the voice output from microphone to byte format
count = read(sData, 0, blockSize);
byte bData[] = short2byte(sData);
WriteToFileAsync.getInstance().writeToFile(bData, 0,
blockSize * bytePerElement);
}
}
和我通过缓冲使用公用缓冲区SDATA FFT的。
and I pass buffer to fft using the common buffer sdata.
sb = ShortBuffer.allocate(blockSize);
sb.put(audioRecorder.sData);
/************ NATIVE DATA/SIGNAL PROCESSING TASK *************/
int pitch = ProcessAudio.process(sb, processed, audioRecorder.count
/ Short.SIZE * Byte.SIZE);
以下是我的C code
Following is my c code
int i;
int j;
short* inBuf = (short*) (*env)->GetDirectBufferAddress(env, inbuf);
double* outBuf = (double*) (*env)->GetDirectBufferAddress(env, outbuf);
int outval = 0;
double temp_sum;
double xcorr[N];
int f = 8000; //8000
int lowr = floor(f / 500);
int upr = ceil(f / 75);
int maxv = 0;
int maxp = 0;
int temp_sum1;
double temp_sum2;
//voice detection
temp_sum2 = 0;
for (i = 0; i < N; i++) {
temp_sum2 = temp_sum2 + (double) inBuf[i] * (double) inBuf[i];
}
if (temp_sum2 > 50000000) { //50000000
// autocorrelation
for (i = 0; i < N; i++) {
temp_sum1 = 0;
for (j = 0; j <= N - i - 1; j++) {
temp_sum1 = temp_sum1 + inBuf[i + j] * inBuf[j];
}
xcorr[i] = temp_sum1;
}
maxv = xcorr[lowr];
maxp = lowr;
for (i = lowr; i <= upr; i++) {
if (xcorr[i] > maxv) {
maxv = xcorr[i];
maxp = i;
}
}
outval = (int) f / maxp;
/***************************** Jian Chen ********************************/
double w[N];
double temp[N];
for (i = 0; i < N; i++) {
w[i] = 0.54 - 0.45 * cos(2 * 3.1415926 * i / N);
}
for (i = 0; i < N; i++) {
temp[i] = ((double) inBuf[i]) * w[i];
}
fftw_plan my_plan;
fftw_complex *in, *out;
/*in = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*2*N);
out = (fftw_complex*) fftw_malloc(sizeof(fftw_complex)*2*N);
my_plan = fftw_plan_dft_1d(2*N, in, out, FFTW_FORWARD, FFTW_ESTIMATE);
*/
in = (fftw_complex*) fftw_malloc(sizeof(fftw_complex) * 16 * N); //2
out = (fftw_complex*) fftw_malloc(sizeof(fftw_complex) * 16 * N); //2
my_plan = fftw_plan_dft_1d(16 * N, in, out, FFTW_FORWARD,
FFTW_ESTIMATE); //2
for (i = 0; i < N; i++) {
in[i][0] = temp[i];
in[i][1] = 0;
}
for (i = N; i < (16 * N); i++) //2*N
{
in[i][0] = 0;
in[i][1] = 0;
}
fftw_execute(my_plan);
double temp1[N];
for (i = 0; i < N; i++) {
temp1[i] = log10(out[i][0] * out[i][0] + out[i][1] * out[i][1]);
if (temp1[i] > 12) {
temp1[i] = 12;
} else if (temp1[i] < 7) {
temp1[i] = 7;
}
outBuf[i] = (temp1[i] * 0.2) - 1.4; //(12.5 6.5;1/6 5/6) (1/6 -1; 12,6)
// overwrite to emphasize the pitch
// *8*4000 now //
if ((i - (int) ((double) outval * (double) 128 / (double) 4000 * 16))
< 4
&& (i
- (int) ((double) outval * (double) 128
/ (double) 4000 * 16)) > 0)
outBuf[i] = 1;
}
fftw_destroy_plan(my_plan);
fftw_free(in);
fftw_free(out);
return outval;
//return temp_sum2;
} else {
for (i = 0; i < N; i++) {
outBuf[i] = 0;
}
return outval = 0;
}
但这种code给我的libc误差致命的信号11 code = 1
任何人都能指出我的错误？
Can anyone point my mistake?
推荐答案
我想，在C code N 是短裤的数量 inBuff 蚂蚁是函数过程的第三个参数。而你为什么使用 audioRecorder.count / Short.SIZE * Byte.SIZE 作为第三个参数。难道不应该只是 audioRecorder.count 。 AudioRecorder的 阅读（）方法返回短裤读，如果你的用户缓冲区哦短裤作为输入参数的数量。
I suppose that in your C code N is the number of shorts in inBuff ant that is the third parameter of the function process. And why do you use audioRecorder.count/ Short.SIZE * Byte.SIZE as third parameter. Shouldn't it be just audioRecorder.count. AudioRecorder's read() method returns the number of shorts read if you user buffer oh shorts as input parameter.
您还可以检查<一href=\"http://stackoverflow.com/questions/17840521/android-fatal-signal-11-sigsegv-at-0x636f7d89-$c$c-1-how-can-it-be-tracked\">this SO质疑来看看行code的给予例外。
You could also check this SO question to see what line of code give the exception.
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