Question

我正努力在Media Foundation中制作卡通着色器，为此，我需要将相机提供的NV12原生转换为RGB24。到目前为止，我对IMFTransform的尝试看起来像这样：设置：

inputVideoTypes = new MFT_REGISTER_TYPE_INFO;
inputVideoTypes->guidMajorType = MFMediaType_Video;
inputVideoTypes->guidSubtype = MFVideoFormat_NV12;
outputVideoTypes = new MFT_REGISTER_TYPE_INFO;
outputVideoTypes->guidMajorType = MFMediaType_Video;
outputVideoTypes->guidSubtype = MFVideoFormat_RGB24;
IMFActivate **transformActivateArray = NULL;
UINT32 MFTcount;
hr = MFTEnumEx(MFT_CATEGORY_VIDEO_PROCESSOR, MFT_ENUM_FLAG_ALL, inputVideoTypes, outputVideoTypes, &transformActivateArray, &MFTcount);
hr = VP->GetAttributes(&VPAttributes);
hr = VPAttributes->SetUINT32(MF_TOPOLOGY_ENABLE_XVP_FOR_PLAYBACK, TRUE);
hr = VP->SetInputType(0, streamType2, 0);
MediaFoundationSamples::LogMediaType(streamType2);
DWORD dwIndex = 4;
hr = VP->GetOutputAvailableType(0, dwIndex, &streamType3);
hr = MFSetAttributeSize(streamType3, MF_MT_FRAME_SIZE, 1280, 720);
hr = streamType3->SetUINT32(MF_MT_FIXED_SIZE_SAMPLES, 1);
hr = MFSetAttributeRatio(streamType3, MF_MT_FRAME_RATE, 30, 1);
hr = MFSetAttributeRatio(streamType3, MF_MT_PIXEL_ASPECT_RATIO, 1, 1);
streamType3->SetUINT32(MF_MT_ALL_SAMPLES_INDEPENDENT, 1);
streamType3->SetUINT32(MF_MT_INTERLACE_MODE, 2);
MediaFoundationSamples::LogMediaType(streamType3);
hr = VP->SetOutputType(0, streamType3, 0);
hr = VP->GetInputStreamInfo(0, &InputInfo);
hr = VP->GetOutputStreamInfo(0, &OutputInfo);

InOnReadSample：

hr = VP->ProcessMessage(MFT_MESSAGE_NOTIFY_BEGIN_STREAMING, NULL);
hr = VP->ProcessInput(0, sample, 0);
DWORD statusFlags;
hr = VP->GetOutputStatus(&statusFlags);
while (statusFlags == 0)
{
    hr = VP->ProcessInput(0, sample, 0);
    hr = VP->GetOutputStatus(&statusFlags);
}
DWORD outputStatus = 0;
IMFSample* outputSample;
MFCreateSample(&outputSample);
MFT_OUTPUT_DATA_BUFFER outputBuffer = {};
outputBuffer.pSample = outputSample;
hr = VP->ProcessOutput(0, 1, &outputBuffer, &outputStatus);

但是问题是ProcessOutput返回hr = E_INVALIDARG，我不知道为什么。奇怪的是OutputInfo和InputInfo。两个dwFlags均为0，但它们的cbSize似乎正常。

MediaTypes的日志：

输入（streamType2）：

MF_MT_FRAME_SIZE    1280 x 720
MF_MT_YUV_MATRIX    2
MF_MT_MAJOR_TYPE    MFMediaType_Video
MF_MT_VIDEO_LIGHTING    3
MF_MT_VIDEO_CHROMA_SITING   1
MF_MT_AM_FORMAT_TYPE    {F72A76A0-EB0A-11D0-ACE4-0000C0CC16BA}
MF_MT_FIXED_SIZE_SAMPLES    1
MF_MT_VIDEO_NOMINAL_RANGE   1
MF_MT_FRAME_RATE    30 x 1
MF_MT_PIXEL_ASPECT_RATIO    1 x 1
MF_MT_ALL_SAMPLES_INDEPENDENT   1
MF_MT_FRAME_RATE_RANGE_MIN  128849018881
MF_MT_VIDEO_PRIMARIES   2
MF_MT_INTERLACE_MODE    2
MF_MT_FRAME_RATE_RANGE_MAX  128849018881
{EA031A62-8BBB-43C5-B5C4-572D2D231C18}  1
MF_MT_SUBTYPE   MFVideoFormat_NV12

输出（streamType3）：

MF_MT_FRAME_SIZE    1280 x 720
MF_MT_MAJOR_TYPE    MFMediaType_Video
MF_MT_FIXED_SIZE_SAMPLES    1
MF_MT_FRAME_RATE    30 x 1
MF_MT_PIXEL_ASPECT_RATIO    1 x 1
MF_MT_ALL_SAMPLES_INDEPENDENT   1
MF_MT_INTERLACE_MODE    2
MF_MT_SUBTYPE   MFVideoFormat_RGB24

有人能告诉我我在做什么错吗？

谢谢！

Answer 1

您尝试在不设置Direct3D感知的情况下转换缓冲区。这对于内存缓冲区很好，在这种模式下，通常应该自己提供输入和输出缓冲区。零OutputInfo.dwFlags恰恰表明了这一点。

因此，MFT_OUTPUT_DATA_BUFFER::pSample初始化在正确的位置上，但是您要提交什么样的示例输出？它是没有连接缓冲液的样品。因此，无效的论点。

使用MFCreateMemoryBuffer为输出的RGB24样本分配内存，然后在ProcessOutput调用中使用它。

E_INVALIDARG，而IMFTransform :: ProcessOutput

1 个答案: