Windows-发出噪音的录音程序
我编写了以下程序，通过windows中的声卡录制声音，并从waveheader的缓冲区打印PCM数据。但它只给出了32600-32700的数据。我的声卡有问题吗？我使用了自动选择源的WAVE_映射器…请帮助我Windows-发出噪音的录音程序,windows,audio,wav,Windows,Audio,Wav,我编写了以下程序，通过windows中的声卡录制声音，并从waveheader的缓冲区打印PCM数据。但它只给出了32600-32700的数据。我的声卡有问题吗？我使用了自动选择源的WAVE_映射器…请帮助我 #include <Windows.h> #include <MMSystem.h> #include <iostream> using namespace std; int main(){ HWAVEIN microHandle; WAVEH
#include <Windows.h>
#include <MMSystem.h>
#include <iostream>
using namespace std;
int main(){
HWAVEIN microHandle;
WAVEHDR waveHeader;
MIXERCAPS mixerCaps;
WAVEFORMATEX format;
//HWAVEOUT hwo; // play
while (1){
const int NUMPTS = 44100 * 0.01; // 10 seconds
int sampleRate = 44100; //can get frequency from here
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
MMRESULT result = 0;
format.wFormatTag = WAVE_FORMAT_PCM; // simple, uncompressed format
format.wBitsPerSample = 8; // 16 for high quality, 8 for telephone-grade
format.nChannels = 1; // 1=mono, 2=stereo
format.nSamplesPerSec = sampleRate; // 22050
format.nAvgBytesPerSec = format.nSamplesPerSec*format.nChannels*format.wBitsPerSample / 8;
// = nSamplesPerSec * n.Channels * wBitsPerSample/8
format.nBlockAlign = format.nChannels*format.wBitsPerSample / 8;
// = n.Channels * wBitsPerSample/8
format.cbSize = 0;
result = waveInOpen(µHandle, WAVE_MAPPER, &format, 0L, 0L, WAVE_FORMAT_DIRECT);
cout << "checking step 1" << endl;
if (result)
{
cout << "Fail step 1" << endl;
cout << result << endl;
Sleep(10000);
return 0;
}
// Set up and prepare header for input
waveHeader.lpData = (LPSTR)waveIn;
waveHeader.dwBufferLength = NUMPTS;// *2;//why *2
waveHeader.dwBytesRecorded = 0;
waveHeader.dwUser = 0L;
waveHeader.dwFlags = 0L;
waveHeader.dwLoops = 0L;
waveInPrepareHeader(microHandle, &waveHeader, sizeof(WAVEHDR));
// Insert a wave input buffer
result = waveInAddBuffer(microHandle, &waveHeader, sizeof(WAVEHDR));
int NumOfMixers = mixerGetNumDevs();
cout << NumOfMixers << endl;
//cout<<(char*)mixerCaps.szPname<<endl;
//system("pause");
cout << "checking step 2" << endl;
if (result)
{
cout << "Fail step 2" << endl;
cout << result << endl;
Sleep(10000);
return 0;
}
//system("pause");
result = waveInStart(microHandle);
cout << "checking step 3......started recording..." << endl;
if (result)
{
cout << "Fail step 3" << endl;
cout << result << endl;
Sleep(10000);
return 0;
}
// Wait until finished recording
do { cout << "still"; } while (waveInUnprepareHeader(microHandle, &waveHeader, sizeof(WAVEHDR)) == WAVERR_STILLPLAYING);
waveInStop(microHandle);
waveInReset(microHandle);
//waveInUnprepareHeader(hwi, lpWaveHdr, sizeof(WAVEHDR));
waveInClose(microHandle);
//printing the buffer
for (int i = 0; i < waveHeader.dwBufferLength; i++)
{
if (waveIn[i] > 0)
cout << i << "\t" << waveIn[i] << endl;
}
}
system("pause");
//waveInClose(microHandle);
return 0;
}
#包括
#包括
#包括
使用名称空间std；
int main（）{
静脉微柄；
波头；
混合器捕捉器混合器捕捉器；
波形格式；
//HWAVEOUT hwo；//播放
而(1){
常量int NUMPTS=44100*0.01；//10秒
int sampleRate=44100；//可以从这里获取频率
short int waveIn[NUMPTS]；/“short int”是一种16位类型；我要求在下面提供16位示例
//对于8位捕获，您将使用“unsigned char”或“BYTE”8位类型
MMRESULT结果=0；
format.wFormatTag=WAVE\u format\u PCM；//简单的未压缩格式
format.wBitsPerSample=8；//16表示高质量，8表示电话等级
format.nChannels=1；//1=单声道，2=立体声
format.nSamplesPerSec=sampleRate；//22050
format.nAvgBytesPerSec=format.nSamplesPerSec*format.nChannels*format.wBitsPerSample/8；
//=nSamplesPerSec*n.通道*wBitsPerSample/8
format.nblockallign=format.nChannels*format.wBitsPerSample/8；
//=n个通道*wBitsPerSample/8
format.cbSize=0；
结果=waveInOpen（µHandle，WAVE_MAPPER，&format，0L，0L，WAVE_format_DIRECT）；
cout一个明显的问题是您混合使用16位和8位。您的缓冲区定义为16位短。请注意您自己的评论：
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
但是，在定义音频格式时，您指定的是8位：
format.wBitsPerSample = 8; // 16 for high quality, 8 for telephone-grade
更改format.wBitsPerSample=16
，或者如果需要8位音频，请执行以下操作：
unsigned char waveIn[NUMPTS];
...
format.wBitsPerSample = 8; // 16 for high quality, 8 for telephone-grade
//printing the buffer
for (int i = 0; i < waveHeader.dwBufferLength; i++)
{
cout << i << "\t" << (int)waveIn[i] << endl;
}
unsigned char waveIn[NUMPTS]；
...
format.wBitsPerSample=8；//16表示高质量，8表示电话等级
//打印缓冲区
对于（int i=0；ishort 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
但是，在定义音频格式时，您指定的是8位：
format.wBitsPerSample = 8; // 16 for high quality, 8 for telephone-grade
更改format.wBitsPerSample=16
，或者如果需要8位音频，请执行以下操作：
unsigned char waveIn[NUMPTS];
...
format.wBitsPerSample = 8; // 16 for high quality, 8 for telephone-grade
//printing the buffer
for (int i = 0; i < waveHeader.dwBufferLength; i++)
{
cout << i << "\t" << (int)waveIn[i] << endl;
}
unsigned char waveIn[NUMPTS]；
...
format.wBitsPerSample=8；//16表示高质量，8表示电话等级
//打印缓冲区
对于（int i=0；i