如何减少音频录制程序(C ++)中的CPU消耗

时间:2018-07-12 06:55:51

标签: c++ cpu cpu-usage audio-recording audio-capture

在互联网的帮助下,我编写了程序,可在麦克风运行时始终捕获其声音。一切正常,但是我需要减少CPU的电量,因为现在大约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);
}

这是产生电荷的整个功能。我该如何减少呢?也许我不能?使用WindowsAPI以外的其他捕获方法?

我试图降低每秒采样数,但这并没有太大帮助。我想这与缓冲区有关,但是需要任何提示。

欢呼

1 个答案:

答案 0 :(得分:0)

正如我们所评论的,cpu消耗是由于do-while循环中的活动等待。我建议使用Event object in windows。 但是,作为即时解决方案,我建议您睡几毫秒(做一些相关的计算,以确切地了解什么是最理想的时间,如@MSalters在其评论中所述)。

基于How do you make a program sleep in C++ on Win 32?,可能的解决方案是:

#include <windows.h>

//....
do {
   Sleep(X);
} 
while (waveInUnprepareHeader(hWaveIn, &waveHeader, sizeof(WAVEHDR)) == WAVERR_STILLPLAYING);
//....