How does OpenCV handle TCP Connections? - sockets

Question

I setup a NetCat Video Stream from my RPi and I am accessing it with OpenCV in the following way:
videoStream = cv2.VideoCapture("tcp://#<my_ip>:<my_port>/")
...
videoStream.release()
Unfortunately I cannot connect to the Stream multiple times without reinitializing it. How does OpenCV tread my tcp connection? Does .release() properly close the socket or what is the right way to close it?

Answer 1

I would comment but I do not have enough points. I had a similar issue. Ultimately, what worked for me is the run netcat with the -k option, which does allow reconnecting:
on RPI:
/opt/vc/bin/raspivid -n -t 0 -w 640 -h 360 -fps 30 -ih -fl -l -o - | /bin/nc -klvp 5000
for nc, the -k option keeps the port listening after the first client disconnects, thereby allowing you to reconnect. You won't need the -v option, it just adds some verbosity.
Another alternative is to
on receiver (Ubuntu, Win10):
nc x.x.x.x 5000 | mplayer -fps 200 -demuxer h264es -
or
gst-launch-1.0 -v tcpclientsrc host=10.60.66.237 port=5000 ! decodebin ! autovideosink
Python code with opencv:
import cv2
cap = cv2.VideoCapture("tcp://10.60.66.237:5000")
while(True):
ret, frame = cap.read()
cv2.imshow('frame', frame)
# the 'q' button is set as the
# quitting button you may use any
# desired button of your choice
if cv2.waitKey(1) & 0xFF == ord('q'):
break
cap.release()
cv2.destroyAllWindows()
Disconnect and reconnect all you want :)