自 Windows Multimedia turned out to be utterly incapable of recording continuous audio ,我得到了使用Windows Core Audio的提示。有一本手册here ,但我不知道如何编写大量的开销代码来使录音工作。谁能提供一个完整的、最小化的连续音频录制到循环缓冲区的实现?
到目前为止,我一直停留在下面的代码中,没有超过 pEnumerator->GetDefaultAudioEndpoint(eRender, eConsole, &pDevice);
行,因为 pEnumerator
仍然是 nullptr
.
#define VC_EXTRALEAN
#define _USE_MATH_DEFINES
#include <Windows.h>
#include <Audioclient.h>
#include <Mmdeviceapi.h>
#define REFTIMES_PER_SEC 10000000
#define REFTIMES_PER_MILLISEC 10000
int main() {
REFERENCE_TIME hnsRequestedDuration = REFTIMES_PER_SEC;
UINT32 bufferFrameCount;
UINT32 numFramesAvailable;
IMMDeviceEnumerator* pEnumerator = NULL;
IMMDevice* pDevice = NULL;
IAudioClient* pAudioClient = NULL;
IAudioCaptureClient* pCaptureClient = NULL;
WAVEFORMATEX* pwfx = NULL;
UINT32 packetLength = 0;
BYTE* pData;
DWORD flags;
CoCreateInstance(__uuidof(MMDeviceEnumerator), NULL, CLSCTX_ALL, __uuidof(IMMDeviceEnumerator), (void**)&pEnumerator);
pEnumerator->GetDefaultAudioEndpoint(eRender, eConsole, &pDevice);
pDevice->Activate(__uuidof(IAudioClient), CLSCTX_ALL, NULL, (void**)&pAudioClient);
pAudioClient->GetMixFormat(&pwfx);
pAudioClient->Initialize(AUDCLNT_SHAREMODE_SHARED, AUDCLNT_STREAMFLAGS_LOOPBACK, hnsRequestedDuration, 0, pwfx, NULL);
pAudioClient->GetBufferSize(&bufferFrameCount); // Get the size of the allocated buffer.
pAudioClient->GetService(__uuidof(IAudioCaptureClient), (void**)&pCaptureClient);
// Calculate the actual duration of the allocated buffer.
REFERENCE_TIME hnsActualDuration = (double)REFTIMES_PER_SEC* bufferFrameCount / pwfx->nSamplesPerSec;
pAudioClient->Start(); // Start recording.
// Each loop fills about half of the shared buffer.
while(true) {
// Sleep for half the buffer duration.
Sleep(hnsActualDuration/REFTIMES_PER_MILLISEC/2);
pCaptureClient->GetNextPacketSize(&packetLength);
while(packetLength != 0) {
// Get the available data in the shared buffer.
pCaptureClient->GetBuffer(&pData, &numFramesAvailable, &flags, NULL, NULL);
if(flags&AUDCLNT_BUFFERFLAGS_SILENT) {
pData = NULL; // Tell CopyData to write silence.
}
// Copy the available capture data to the audio sink.
//hr = pMySink->CopyData(pData, numFramesAvailable, &bDone);
pCaptureClient->ReleaseBuffer(numFramesAvailable);
pCaptureClient->GetNextPacketSize(&packetLength);
}
}
pAudioClient->Stop();
return 0;
}
编辑(2021 年 7 月 24 日):
这是我的代码更新,用于故障排除:
#define VC_EXTRALEAN
#define _USE_MATH_DEFINES
#include <Windows.h>
#include <Audioclient.h>
#include <Mmdeviceapi.h>
#include <chrono>
class Clock {
private:
typedef chrono::high_resolution_clock clock;
chrono::time_point<clock> t;
public:
Clock() { start(); }
void start() { t = clock::now(); }
double stop() const { return chrono::duration_cast<chrono::duration<double>>(clock::now()-t).count(); }
};
const uint base = 4096;
const uint sample_rate = 48000; // must be supported by microphone
const uint sample_size = 1*base; // must be a power of 2
const uint bandwidth = 5000; // must be <= sample_rate/2
float* wave = new float[sample_size]; // circular buffer
void fill(float* const wave, const float* const buffer, int offset) {
for(int i=sample_size; i>=offset; i--) {
wave[i] = wave[i-offset];
}
for(int i=0; i<offset; i++) {
const uint p = offset-1-i;
wave[i] = 0.5f*(buffer[2*p]+buffer[2*p+1]); // left and right channels
}
}
int main() {
for(uint i=0; i<sample_size; i++) wave[i] = 0.0f;
Clock clock;
#define REFTIMES_PER_SEC 10000000
#define REFTIMES_PER_MILLISEC 10000
REFERENCE_TIME hnsRequestedDuration = REFTIMES_PER_SEC;
UINT32 bufferFrameCount;
UINT32 numFramesAvailable;
IMMDeviceEnumerator* pEnumerator = NULL;
IMMDevice* pDevice = NULL;
IAudioClient* pAudioClient = NULL;
IAudioCaptureClient* pCaptureClient = NULL;
WAVEFORMATEX* pwfx = NULL;
UINT32 packetLength = 0;
BYTE* pData;
DWORD flags;
CoInitializeEx(NULL, COINIT_MULTITHREADED);
CoCreateInstance(__uuidof(MMDeviceEnumerator), NULL, CLSCTX_ALL, __uuidof(IMMDeviceEnumerator), (void**)&pEnumerator);
pEnumerator->GetDefaultAudioEndpoint(eRender, eConsole, &pDevice);
pDevice->Activate(__uuidof(IAudioClient), CLSCTX_ALL, NULL, (void**)&pAudioClient);
pAudioClient->GetMixFormat(&pwfx);
println(pwfx->wFormatTag);// 65534
println(WAVE_FORMAT_PCM);// 1
println(pwfx->nChannels);// 2
println((uint)pwfx->nSamplesPerSec);// 48000
println(pwfx->wBitsPerSample);// 32
println(pwfx->nBlockAlign);// 8
println(pwfx->wBitsPerSample*pwfx->nChannels/8);// 8
println((uint)pwfx->nAvgBytesPerSec);// 384000
println((uint)(pwfx->nBlockAlign*pwfx->nSamplesPerSec*pwfx->nChannels));// 768000
println(pwfx->cbSize);// 22
pAudioClient->Initialize(AUDCLNT_SHAREMODE_SHARED, AUDCLNT_STREAMFLAGS_LOOPBACK, hnsRequestedDuration, 0, pwfx, NULL);
pAudioClient->GetBufferSize(&bufferFrameCount); // Get the size of the allocated buffer.
pAudioClient->GetService(__uuidof(IAudioCaptureClient), (void**)&pCaptureClient);
// Calculate the actual duration of the allocated buffer.
//REFERENCE_TIME hnsActualDuration = (double)REFTIMES_PER_SEC* bufferFrameCount / pwfx->nSamplesPerSec;
pAudioClient->Start(); // Start recording.
while(running) {
pCaptureClient->GetNextPacketSize(&packetLength); // packetLength and numFramesAvailable are either 0 or 480
pCaptureClient->GetBuffer(&pData, &numFramesAvailable, &flags, NULL, NULL);
const int offset = (uint)numFramesAvailable;
if(offset>0) {
fill(wave, (float*)pData, offset); // here I add pData to the circular buffer "wave"
}
while(packetLength != 0) {
pCaptureClient->GetBuffer(&pData, &numFramesAvailable, &flags, NULL, NULL); // Get the available data in the shared buffer.
if(flags&AUDCLNT_BUFFERFLAGS_SILENT) {
pData = NULL; // Tell CopyData to write silence.
}
pCaptureClient->ReleaseBuffer(numFramesAvailable);
pCaptureClient->GetNextPacketSize(&packetLength);
}
sleep(1.0/120.0-clock.stop());
clock.start();
}
pAudioClient->Stop();
}
最佳答案
你没有调用 CoInitializeEx
,因此所有 COM 调用都会失败。
您还应该测试所有调用以查看它们是否返回错误。
解决评论中提出的问题:
我相信,如果您想在共享模式下操作端点,那么您必须使用GetFixFormat
返回的参数。这意味着:
您只能使用一个采样率(除非您编写代码来执行转换,这是一项艰巨的任务)
如果您希望样本为 float ,则必须自行转换它们
要编写在所有机器上运行的代码,您必须满足混合格式向您抛出的任何问题。这可能是:
16 位整数
24 位整数 (nBlockAlign = 3)
32 位容器中的 24 位整数 (nBlockAlign = 4)
32 位整数
32 位浮点(罕见)
64 位浮点(根据我的经验,这是闻所未闻的)
样本将按照代码运行所在机器的 native 字节顺序排列,并且是交错的。
因此,请列出 pwfx
中的各种参数,并为您想要支持的每种示例格式编写相关代码。
假设您希望将 float
标准化为 -1 .. +1
和 2 channel 输入数据,您可以对 16 位整数执行此操作,例如:
const int16_t *inbuf = (const int16_t *) pData;
float *outbuf = ...;
for (int i = 0; i < numFramesAvailable * 2; ++i)
{
int16_t sample = *inbuf++;
*outbuf++ = (float) (sample * (1.0 / 32767));
}
请注意,我通过乘以倒数来避免(缓慢的)浮点除法(编译器将预先计算 1.0/32767
)。
剩下的就交给你了。
关于c++ - 如何在 Windows 10 上使用 C++ 将连续的原始音频数据记录到循环缓冲区中?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/63456223/