C ++ OpenCV获取编码的网络摄像头流

Question

I am currently work on a project that capture video from webcam and send the encoded stream via UDP to do a real time streaming.

#include "opencv2/highgui/highgui.hpp"
#include <iostream>

using namespace cv;
using namespace std;

int main(int argc, char* argv[])
{
VideoCapture cap(0); // open the video camera no. 0

double dWidth = cap.get(CV_CAP_PROP_FRAME_WIDTH); //get the width of frames of the video
double dHeight = cap.get(CV_CAP_PROP_FRAME_HEIGHT); //get the height of frames of the video

while (1)
{
    Mat frame;

    bool bSuccess = cap.read(frame); // read a new frame from video

if (!bSuccess) //if not success, break loop
{
cout << "Cannot read a frame from video stream" << endl;
break;
}
return 0;
}

Some people say that the frame get from cap.read(frame) is already the decoded frame,I have no idea how and when does that happen. And what I want is the encoded frame or stream. What should I do to get it? Should I encoded it back again?

Answer 1

According to the docs , calling VideoCapture::read() is equivalent to calling VideoCapture::grab() then VideoCapture::retrieve() .

The docs for the Retrieve function say it does indeed decode the frame.

Why not just use the decoded frame; presumably you'd be decoding it at the far end in any case?

Answer 2

OpenCV API does not give access to the encoded frames.

You will have to use a more low-level library, probably device and platform dependent. If your OS is Linux, Video4Linux2 may be an option, there must be equivalent libraries for Windows/MacOS. You may also have a look at mjpg-streamer , which does something very similar to what you want to achieve (on linux only).

Note that the exact encoding of the image will depend on your webcam, some usb webcam support mjpeg compression (or even h264), but other are only able to send raw data (usually in yuv colorspace).

Another option is to grab the decoded image wit Opencv, and reencode it, for example with imencode . It has the advantages of simplicity and portability, but image reencoding will use more resource.