cv::Mat* is pointer to the cv::Mat. Believing that much, the trivial assignments should work:
cv::Mat x;
cv::Mat *xp = &x;
cv::Mat y = *xp;
Beware simple assignment does not copy the buffer, so you must be sure that Mat object referenced by that pointer lives at least as long as you y does. If you are not sure, create a complete copy of it:
cv::Mat y = xp->clone();
It is really strange to see pointers to cv::Mat, normally should not be required. I would not recommend, but when working with the third party code it may be no choices.
Asked: 2019-07-16 01:16:05 -0600
Seen: 1,592 times
Last updated: Jan 27 '20
WHY is there a
cv::Mat*in your code (it shouldn't be !) ?please show us, so we get a better idea, what to do.
also please look up "how to dereference c++ pointers", it's a very basic c++ topic, and you should have it covered before venturing into more complicated things, like computer-vision.
RED FLAG here. unless it's open src you wouldn't know, whether to delete it in the end or not.
again, please show us some code (function signatures, etc)
Thanks @berak,
Actually it is not in my code, Third party sdk is response their result as cv::Mat* .
According by their document, OpenCV can be used in case if OpenCV library is available.
I will check with third party vendor again. thanks again.
5 posts already, and we still don't know, which sdk you're talking about (link ?) what
result.image.mat();might be, exactly, constraints of it's usage (null pointers ? pointer ownage ?), etc.https://www.vision-components.com/en/...
https://www.carrida-technologies.com/...
Well that's remarkably un-informative. I think it's just returning a pointer to the internal cv::Mat, so don't delete it. But it copies a cv::Mat you use to construct it, with an optional bool.
Oh dear, it actually stores a cv::Mat* as it's member variable. Eww.