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Why mat.ptr<>() has the same address among all the elements inside std::vector<cv::Mat>?

asked 2017-11-25 03:55:28 -0500

updated 2017-11-25 03:58:49 -0500

If I construct outfea_ in the following manner, then you'll find the three output have exactly the same value. Is it a bug in opencv?

std::vector<cv::Mat> outfea_(3, cv::Mat(5, 5, CV_32FC1, cv::Scalar(0)));
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answered 2017-11-25 04:14:14 -0500

berak gravatar image

updated 2017-11-25 04:22:12 -0500

this is, because all the Mat's in your vector really point to the same data !

the vector constructor makes a "shallow" copy of the Mat argument only, so they all share the same data ptr.

(this is not restricted to cv::Mat, you'll see that with anything, that contains a pointer, the vector makes shallow copies of the pointer, not whatever it points to)

the only remedy is to fill the vector with independant Mat's like:

// c++11
std::vector<cv::Mat> outfea_ = {
 cv::Mat(5, 5, CV_32FC1, cv::Scalar(0)),
 cv::Mat(5, 5, CV_32FC1, cv::Scalar(0)),
 cv::Mat(5, 5, CV_32FC1, cv::Scalar(0))

 // c++98
std::vector<cv::Mat> outfea_; 
for (int i=0;i<3; i++) {
 outfea_.push_back(cv::Mat(5, 5, CV_32FC1, cv::Scalar(0)));
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Oh I got it, thanks a lot!

charlesjiangxm gravatar imagecharlesjiangxm ( 2017-11-25 05:12:16 -0500 )edit

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Asked: 2017-11-25 03:55:28 -0500

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Last updated: Nov 25 '17