How to create cv::Mat from buffer (array of T* data) using a template function?

Question

I'd like to write a template function to copy data referenced by pointer T* image to cv::Mat. I am confusing how to generalize T and cv_type matching.

template<typename T>
cv::Mat convert_mat(T *image, int rows, int cols) {
    // Here we need to match T to cv_types like CV_32F, CV_8U and etc.
    // The key point is how to connect these two
    cv::Mat mat(rows, cols, cv_types, image);
    return mat;
}

I am new to template programming, I am quite confused how to implement T-cv_types correspondence.

Anyone has any idea? Thank you!!!

I don't really understand the question - what logic do you use for mapping types to "cv_types" when you don't use templates? — UnholySheep
– UnholySheep, Commented Jan 11, 2018 at 13:10
you can probably try it with cv::Mat_<T> which does not require the cv_type since it is templated — api55
– api55, Commented Jan 11, 2018 at 13:10

Kinght 金 · Accepted Answer · 2018-09-03 08:30:51Z

17

Use cv::DataType<T>::type .

Here is an example.

// Create Mat from buffer 
#include <opencv2/opencv.hpp>
using namespace std;
using namespace cv;


/*
//! First version
//! 2018.01.11 21:16:32 (+0800)
template <typename T>
Mat createMat(T* data, int rows, int cols) {
    // Create Mat from buffer
    Mat mat(rows, cols, cv::DataType<T>::type);
    memcpy(mat.data, data, rows*cols * sizeof(T));
    return mat;
}
*/

//! Second version 
//! 2018.09.03 16:00:01 (+0800) 
template <typename T>
cv::Mat createMat(T* data, int rows, int cols, int chs = 1) {
    // Create Mat from buffer 
    cv::Mat mat(rows, cols, CV_MAKETYPE(cv::DataType<T>::type, chs));
    memcpy(mat.data, data, rows*cols*chs * sizeof(T));
    return mat;
}

int main(){
    int    arr1[4] = {1,2,3,4};
    double arr2[4] = {1.1,2.2,3.3,4.4};

    Mat mat1 = createMat<int>(arr1, 2,2);
    Mat mat2 = createMat<double>(arr2, 2,2);
    cout << "Mat1:\n"<< mat1 <<endl;
    cout << "Mat2:\n"<< mat2 <<endl;
}

Result:

Mat1:
[1, 2;
 3, 4]
Mat2:
[1.1, 2.2;
 3.3, 4.4]

edited Sep 3, 2018 at 8:30

answered Jan 11, 2018 at 13:11

Kinght 金

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1 Comment

rayryeng Over a year ago

Nice answer. I've bookmarked this. I've always wanted a streamlined way of doing this. Thank you!

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