在互联网的帮助下,我编写了一个程序,它在运行时一直从麦克风捕获音频。一切都很好,但我需要减少 CPU 的费用,因为现在它大约是 30-35%。
int SoundCapture()
{
const int NUMPTS = 8000 * 1; // Sample rate * seconds
int sampleRate = 8000;
short int waveIn[NUMPTS]; // 'short int' is a 16-bit type; I request 16-bit samples below
// for 8-bit capture, you'd use 'unsigned char' or 'BYTE' 8-bit types
HWAVEIN hWaveIn;
WAVEFORMATEX waveform;
WAVEHDR waveHeader;
waveform.wFormatTag = WAVE_FORMAT_PCM; // simple, uncompressed format
waveform.nChannels = 1; // 1=mono, 2=stereo
waveform.nSamplesPerSec = 8000;
waveform.nAvgBytesPerSec = 8000; // = nSamplesPerSec * n.Channels * wBitsPerSample/8
waveform.nBlockAlign = 1; // = n.Channels * wBitsPerSample/8
waveform.wBitsPerSample = 8; // 16 for high quality, 8 for telephone-grade
waveform.cbSize = 0;
MMRESULT result = waveInOpen(&hWaveIn, WAVE_MAPPER, &waveform, 0, 0, WAVE_FORMAT_DIRECT);
if (result)
{
std::cout << "Something wrong with WaveOpen";
std::cin.ignore(2);
return 0;
}
// Set up and prepare header for input
waveHeader.lpData = (LPSTR)waveIn; //pointer to waveform buffer
waveHeader.dwBufferLength = NUMPTS;
waveHeader.dwBytesRecorded = 0;
waveHeader.dwUser = 0L;
waveHeader.dwFlags = 0L;
waveHeader.dwLoops = 0L;
waveInPrepareHeader(hWaveIn, &waveHeader, sizeof(WAVEHDR));
// Insert a wave input buffer
result = waveInAddBuffer(hWaveIn, &waveHeader, sizeof(WAVEHDR));
if (result)
{
std::cout << "Something wrong with waveInAddBuffer";
std::cin.ignore(2);
return 0;
}
// Commence sampling input
time_t t = time(0); // get time now
result = waveInStart(hWaveIn);
if (result)
{
std::cout << "Something wrong with WaveStart";
std::cin.ignore(2);
return 0;
}
// Wait until finished recording
do {} while (waveInUnprepareHeader(hWaveIn, &waveHeader, sizeof(WAVEHDR)) == WAVERR_STILLPLAYING);
SaveWavFile(&waveHeader, t);
waveInClose(hWaveIn);
}
这是产生电荷的全部功能。我怎样才能减少它?或者我不能?使用 WindowsAPI 以外的其他捕获方法?
我试图减少每秒的样本,但这并没有太大帮助。我想这与缓冲区有关,但需要任何提示。
干杯